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Probability Tricks & Tips

INTRODUCTION:

Today I came with one of the easiest topic in Aptitude Section that is PROBABILITY.This topic is asked for 1 Mark or 2 Mark in Prelims Examination & the Problem from this section will be quite easy & we can solve in a minute.

To Crack the Exam the first thing we should know before knowing how to solve the question is selection of the question.It is important to choose the question which can be solved easily.So,Probability is that kind of Section in which we can solve the question in few seconds.It is wise to attempt the questions from Probability because they can’t trick you after a certain extent.

There is only 1 Rule in Probability that Number of Possible ways/Total Number of Sample Space

1.Based on Coins

2.Based on Dice

3.Based on Cards

#.1 TYPE 1:

PROBLEMS BASED ON COINS:

The Possible Number of Outcomes is always Calculated by 2∧n

When a Coin is Tossed then the Possible ways=2¹=2(Head or Tail)

When Three Coins are Tossed then the Possible ways=2³=8(HHH,HHT,HTH,HTT,THH,THT,TTH,TTT)

The Question from this Coin Pattern May have 3 words to trick you that is

ATMOST

ATLEAST

EXACTLY

If in the Question they specify Atmost 2 heads then they are saying that Maximum of two heads

If they Specify Atleast 2 heads then they are saying there should Minimum of Two heads There May be more than 2 heads

If they Specify Exactly 2 heads then there should be only 2 heads not more than that not less than that.

#.2.TYPE 2:

PROBLEMS BASED ON DICE:

The same Logic Applies here that is the Possible number of Outcomes is 2 Power n

When a Single Dice is thrown the number of Possibility=6¹=6(1,2,3,4,5,6)

When two Dice are thrown then the number of Possible Outcomes=6²=36

[(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)

(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)

(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)

(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)

(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)

(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)]

#.3.TYPE 3

PROBLEMS BASED ON CARDS:

In Card Based Problem Remember this Hierarchy

13 Cards in Each Category is:

Numbers (1 to 9) 9 cards

Ace-1,Jack-1 Queen-1 King-1 (4 cards)

Let Us Look Into Problems From Each Type:

1)What is the Probability of getting Exactly 1 Tail when Three Dice are thrown?

EXPLANATION

The Total number of Possibility=2³=8(HHH,HHT,HTH,HTT,THH,THT,TTH,TTT)

So getting Exactly One Tail={(HHT),(HTH),(THH)}

Probability=Possible Outcomes/Total number of Possibility

Probability=3/8

2)When two dice are thrown at a random Find what is the Probability of Same number in each dice?

EXPLANATION

Same Dice Number of events={(1,1)(2,2)(3,3)(4,4)(5,5)(6,6)}=6

Total Events=36

Probability=6/36=1/6 ways

3)What is the Probability of getting sum greater than 6 when two dice are thrown?

EXPLANATION

Sum greater than 6 when two dice are={(1,6)(2,5)(2,6)(3,4)(3,5)(3,6)(4,3)(4,4)(4,5)(4,6)(5,2)(5,3)(5,4)(5,5)(5,6)(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)}

=21/36=7/12 ways

4)When 2 cards are drawn at Random,what is the Probability that Both the cards are Ace?

EXPLANATION

Both the Card are Ace:

Total number of ways=52C2=52*51/1*2=26*51

Number of Possible Outcomes:

Total number of Ace =1*4=4 cards

From this 2 cards has to be drawn out=4C2=4*3/1*2=6

Therefore,

Possible Outcomes=6/26*51=1/221

TOPICS TO BE DISCUSSED

1)BASED ON CARDS

2)BASED ON  BALLS

#.1.TYPE 1:

BASED ON CARDS:

Friends in the last Article we had discussed problems on cards with the Keyword Both.In this Article we will discuss problems based on the Keyword

No, One from one type & Another from another type

13 Cards In Diamond,Heart,Spade & Club Contains:

Numbered Card:2,3,4,5,6,7,8,9,10(Totally 9  Numbered Cards)

Face Card:There are Face Card they are (ACE,KING QUEEN & JACK)

9+4=13 CARDS in each Suit.

Now let us look into the Problem.

1)When 1 card is drawn at Random,What is the Probability that there will be NO King Card?

EXPLANATION

Probability=Number of Possible Ways/Total Number of Events

Here we are going to draw 1 card therefore the Total number of Events or Sample Space=52C1=52

There are 4 King Cards in 52 Cards

1 from Diamond,1 from Heart,1 from club,1 from Spade,Totally 4 King Cards  in 52 Cards.

=52-4=48

Therefore Possible ways=48C1

Probability of getting No King:

=48/52

=12/13

Friends Shall We Solve Together,

3 Cards are drawn,Total Possibility=52C3=52*51*50/1*2*3

=26*17*50

Now,let us find the Possible Number of ways of getting 1 Ace,1 King & 1 Jack

There are 4 Ace in 52 Cards,

1 King in 52 Cards,

1 Jack in 52 Cards,

Now,we have to choose 1 card from each face.

4C1 * 4C1 * 4C1=4*4*4

Therefore,

=4*4*4/ 26*17*50

Probability of getting 3 Cards=16/5525

#.2.TYPE 2:

BASED ON  BALLS:

In this type,Questions will be asked in such a way that certain balls are drawn in different colours and after that there will be two conditions after this

1)They were Replaced.

2)They were not Replaced.

1)THEY WERE REPLACED:

A bag contains 8 Red & 10 Black Balls.If two draws of three balls are drawn randomly & the balls are replacedafter the first draw then what is the  probability that three  balls are Red Balls at the first draw and  Black Balls at the Second Draw?

EXPLANATION

A Bag Contains 8 Red & 10 Black Balls

At First Draw:

They will draw three balls of Red Colour,

Number of Possible ways=8C3(Since 3 Red Balls are to be drawn from 8 Red Balls)

Total Number of Ways=18C3(Since 3 Red Balls are to be drawn from 18 Balls)

We all know the formula to find the Probability,

=Number of Possible Ways/Total Number of Ways

=8C3/18C3

8C3=8*7*6/1*2*3=56

18C3=18*17*16/1*2*3=816

=56/816

Probability of the First Draw=7/102

At Second Draw:

We want to draw 3 Balls from 10 Black Balls

Possible Ways=10C3=10*9*8/1*2*3=120

Total Number of Ways=18C3=18*17*16/1*2*3=816

Probability=120/816=5/34

Probability of the Second Draw:5/34

Total Probability=7/102 * 5/34

We had Used * here Do You know why?

Because In the Question they ask us to find the Probability of  First Draw & Second Draw

And is Used So we have to Multiply it,

If they said to Find First Draw or Second Draw then We would add it .

Note:If “AND” is represented in the question then we have to “Multiply” the term.

If “OR” is represented in the question then we have to “Add” the term.

2)Without Replacement:

A Bag Contains 8 Red Balls & 10 Black Balls.Four Balls are Drawn one by one & they are not replaced.Find the Probability if the balls drawn are of Alternate of different Colours?(That is,if 1st ball is Red 2nd would be Black …,Vice Versa)

EXPLANATION

Let us Assume if 1st Ball Drawn of Red Colour then,

Probability of 1st Ball:

Possible Ways=8C1=8

Total Ways=18C1=18

Probability of First Ball=8/18=4/9

Since the Ball is not Replaced,now there will be 7Red Balls & Black Balls Remain the Same Totally 17 Balls

Probability of 2nd Ball(That is Black):

Possible Ways=10C1=10

Total Ways=17C1=17

Probability of Second Ball=10/17

This Ball will also be not Replaced then there will be 7Red Balls 9 Black Balls & the Total number of Balls=16 balls

Probability of Third Ball:(Red)

Possible Ways=7C1=7

Total Ways=16C1=16

Probability of Third Ball=7/16

This Ball also will not be replaced then there will be 6 Red balls & 9 Black Balls & the total number of Balls=15 Balls

Probability of Fourth Ball(Black)

Possible Ways=9C1=9

Total Ways=15C1=15

Probability of Fourth Ball=9/15=3/5

Since we want all the Four ball(AND) Therefore we must Multiply individual Probability

4/9 * 10/17 * 7/16 * 2/5

Probability of drawing 4 balls alternatively( if we start with Red Ball)=7/102

First Ball(Black)=10/18=5/9

Second Ball(Red)=8/17

Third Ball(Black)=9/16

Fourth Ball(Red)=7/15

When Four Balls are drawn the Probability is=7/102

Now the Probability is First Ball May be RED or Second Ball May be BLACK

Now we have to add because the Term OR is used

7/102 + 7/102 =14/102=7/51

Probability of an event happening = Concerned Events / Total Events

Probability of an event happening is denoted by P(E)
Probability of an event not happening is denoted by P(Ē).

AndP(E) + P(Ē) = 1

Types of Events:
1. If ‘And’ event is given then we multiply or count events together.
2. If ‘Or’ event is given then we ‘add’ the two or more events.

Classical Cases:
The classical cases discussed in probability are based on following points:

1. Dice:

The dice used here is the one used to play ‘Ludo’. A typical dice has numbers 1, 2, 3, 4, 5 and 6 are written over its six faces as shown below.

When a dice is thrown the number that appears on upper face is the concerned event. The questions based on dice are mainly of two types(not exhaustive):

(I) When Only One dice is thrown once:
In such cases, the number rolling on playing a dice is either 1 or 2 or 3 or 4 or 5 or 6.
Here, the concerned event of rolling out ‘1’ is 1 only because 1 is written only one face. And, total events = 6 because total different numbers written different faces are six in total.
So, Probability of rolling number 1 = 1/6

Similarly, Probability of rolling number 2 ( or any number from 3 to 6) = 1/6

What is the probability of getting an even number on rolling a dice?

Now, concerned event should have an even number which are 2, 4 and 6
So, Concerned Event = 3
Total Event = 6
So, Probability = 3/6 = 1/2

(II) When Two dices are thrown:
In such cases, either of the two things happen either two dices are thrown simultaneously and the numbers appearing on top faces of both dices are noted and summed up; or one dice is thrown two times in a row and the numbers appearing on the top faces in the two times are noted and summed up. Whatever is done, the treatment is same in either the cases, two dices at once or rolling one dice twice. So, this summed up number is the concerned event in such question.

The various combinations of numbers that can turn up on throwing two dices (or one dice twice) can be listed as below –

For example: (1, 6) shows that ‘1’ would turn up on dice 1 and ‘6’ would turn up on dice 2. Here, total outcome is the total number of combinations stated above: (1,1,); (1,2) ……. (6,5); (6,6) = 36

The questions asked can be of following type:

Qs. 1 – What is the probability of getting a combination of ‘5’ and ‘3’ on throwing two dices?

Solutions – Now the concerned event should have number ‘5’ and ‘3’ so concerned events = 2{(3,5) and (5,3)}
Total events = 36
⇒ Probability = 2/36 = 1/18

Qs. 2 –  What is the probability of getting a sum of ‘10’ on rolling a dice twice?

Solutions – Now, the concerned events should have sum of 10 i.e. No. on dice 1 +No. on dice 2 = 10
This can be seen in these cases: (4,6); (5,5) and (6,4)
So, Concerned Events = 3
And, we know Total Events are always ‘36’
So, Probability  = 3/36 = 1/12

2. Coins:

Coin is a currency token which has two faces, one is head and other is tail. So, when throw a coin in air and when it lands it might have either a head or tail. Coin questions can be three types as shown below:

I. One Coin once:
When a coin is tossed is only once then there can be two outcomes either a head or a tail. In such cases, total events = 2
Question: What is the probability of getting a head in a toss?
Solutions – Concerned event = 1(One head)
Total Event = 2
⇒ P(E) = 1/2

II. Two Coins or One Coin Twice:
When two coins are tossed together or one coin is tossed in twice then following outcomes can be obtained:

Here, ‘H’ = Head; ‘T’ = Tail.

(H,T) shows that on coin 1 it’s Head while on coin 2 it’s Tail. Here, we can see that Total Event = 4

Question: What is the Probability of getting at most one head on tossing a coin?
Solutions – At most one head means there can be 0 head or there can be 1 head.
So, Concerned Event = 3 {(H,T) (T,H) (T,T)}
Total Events = 4
⇒ P(E) = 3/4

III. Three Coins or One Coin Thrice:

When three coins are tossed together or one coin is tossed in thrice then following outcomes can be obtained:
Here, ‘H’ = Head; ‘T’ = Tail.
(HHH), (HHT), (HTH), (THH), (HTT), (THT), (TTH), (TTT)
Here, (HTH) shows that coin 1 has a head, coin 2 has a Tail while coin 3 has a head.
In such cases, Total Events = 8

Question: Three unbiased coins are tossed. What is the probability of getting at least 2 heads?
So, Concerned Events = 4 {(HHH), (HHT), (HTH), (THH)}
Total Events = 8
⇒ P(E) =4/8 =1/2

3. Cards:

There are four kinds of symbol used in playing cards. The etymology for different symbols is as below:
i) Spade ⇒ ♠ ⇒ Black in color (13 in number)
ii) Club ⇒ ♣ ⇒ Black in color (13 in number)
iii) Heart ⇒ ♥ ⇒ Red in color (13 in number)
iv) Diamonds ⇒ ♦ ⇒ Red in color (13 in number)

Each of these 4 variants have 13 numbers each as 1, 2, 3 …. 10 and, Jack, Queen, King and Ace. There are
1) 26 red cards and 26 black cards.
2) 4 cards each of 1, 2, 3 …. 10 and, Jack, Queen, King and Ace.
3) 13 cards each of Spade, Heart, Club and Diamond.
So, in total there are 13 × 4 = 52 cards.

1. One card drawn:
In such types of question a card is drawn from the pack of cards. Here, the Total Events = 52

Question: What is the probability of getting a King of Spade or Queen of Heart in one draw?

Solutions – Here, ‘King of Spade or Queen of Heart’ means that either the card can be the Spade King or Heart Queen. Clearly, there is only one King of Space and only one Queen of Heart.
So, Concerned Event = 2
Total Event = 52
⇒ P(E) = 2/52 = 1/26

2. More than One Card drawn:
In such questions when more than One card is drawn we use the concept of Combination formula. For example the question below:

Question – Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart is ____ ?
Here, the ‘one spade’ card has to be drawn from 13 spade cards, so its event = 13C1
And, ‘one heart’ card has to be drawn from 13 heart cards, so its event  = 13C1
So, Concerned Event = 13C1×13C1
Total Events (as two cards are to be drawn from 52) = 52C2
So, P(E) = (13C1×13C1)÷ 52C2  =  (13 x 13) x 2 / 52 x 51 = 13 / 102

Question –  Two cards are drawn together from a pack of 52 cards. The probability that either both are red or both are Kings ____ ?
Here, the ‘both red’ cards have to be drawn from 26 red cards, so its event = 26C2
Or, ‘both king’ cards have to be drawn from 4 King cards, so its event  = 4C2
But there are Two Red Kings which are common in both Red cards & King cards so they have been double counted. And, they will be taken out of two red king cards only so they’ll be deducted from concerned events.
So, Concerned Event = 26C2  + 4C– 2C2
Total Events (as two cards are to be drawn from 52) = 52C2
So, P(E) = (26C2  + 4C– 2C2)÷ 52C2  = [ (26 x 25) + (4 x 3) – 1 ] / 52 x 51 = 330 / 1326 = 55 / 221

4. Balls:

In such questions, a bag contains certain balls and some ball(s) is(are) drawn.

I. One ball drawn:

Question  – In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked at random. What is the probability that it is neither red nor green?

Solutions – Total Events = 8 (red) + 7 (blue) + 6 (green) = 21
Since, the selected ball has to be neither red nor green then it’d be Blue and blue balls are 7.

So, Concerned Event = 7 ⇒ P(E) = 7/21 = 1/3

II. More than One ball drawn without replacement:

Question – A box contains 10 black or 10 white balls. The probability of drawing two balls of same colors?

Solution – Total Events (as 2 balls are drawn) = 20C2
The balls drawn can either be both black color or both white color and for ‘OR’ event we add up the two events.
So, Concerned Event = 10C2 (if black) + 10C2 (if white)
⇒ P(E) = (10C2 + 10C2)÷ 20C= 10 x 9 + 10 x 9 / 20 x 19 = 9 / 19

Question – A box contains 10 black and 10 white balls. The probability of drawing two balls of same colors?

Solution – Total Events (as 2 balls are drawn) = 20C2
The balls drawn can be both black color or both white color. for ‘And’ event we multiply the two events.

First case – If both ball is white –

So, Concerned Event = 10C0 (if black) x 10C2 (if white)

⇒ P(E) = (10C0 x 10C2)÷ 20C= 1 x 10 x 9 / 20 x 19 = 9 / 38

Second case – If both ball is black

So, Concerned Event = 10C2 (if black) x 10C0 (if white)

⇒ P(E) = (10C2 x 10C0)÷ 20C= 10 x 9 x 1 / 20 x 19 = 9 / 38

Other Miscellaneous questions:

Qs 1. Four persons are chosen at random from a group of 3 men, 2 women and 4 children. What is the probability of exactly two of them being children?

Solution – Total People = 3 + 2 + 4 = 9
Since, 4 people are chosen from 9 so Total Event = 9C4
And, since 2 of the chosen people have to be children so these 2 person have to be from 4 children so Concerned Event for children= 4C2
And, other 2 people will be from Men & Women (3+2 = 5), their Concerned event = 5C2
⇒ Concerned Event (for all 4 people) = 4C×5C2
⇒ P(E) = (4C×5C2) ÷ 9C4 = (4 x 3 x 5 x 4) x (2 x3) / 9 x 8 x 7 x 6 = 10 / 21

Qs. 2. A and B give exam. Chance of husband’s selection is 1/7 and of wife’s selection is 1/5. Find probability of only one of them is selected.

Solution – P( husband selecting) = 1/7
P (husband not selecting) = 1 – 1/7 = 6/7
P( wife selecting) = 1/5
P (wife not selecting) = 1 – 1/5 = 4/5
Probability of only one of them is selected = P( husband selecting)× P (wife not selecting) + P( wife selecting)× P (husband not selecting) =  (1/7 x 4/5) + (1/5 x 6/7) = 10/35 = 2/7.

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Probability: A mathematically measure of uncertainty is known as probability.
Random Experiment: An experiment in which all possible outcomes are known and exact
Outcome can be not be predicted, is called a random experiment.
Eg. Rolling an unbiased dice has all six outcomes (1, 2, 3, 4, 5, 6 ) known but exact outcome can be predicted.
Outcome: The result of a random experiment is called an outcome.
Sample Space: The set of all possible outcomes of a random experiment is known as sample space.
eg . The sample space in throwing of a dice is the set (1, 2, 3, 4, 5, 6)
Trial : The performance of a random experiment is called a trial.
eg. The tossing of a coin is called trial
Event: An event is a set of experimental outcomes, or in other words it is a subset of sample space.
eg. On tossing of a dice, let A denotes the event of even number appears on top A: { 2, 4, 6 }
Mutually Exclusive Events : Two or more events are said to be mutually exclusive if the occurrence of any one excludes the happening of other in the same experiment.
eg. On tossing of a coin it head occur, then it prevents happing of tail, in the same single experiment.
Exhaustive Events : All possible outcomes of an event are known as exhaustive events.
eg. In a through of single dice the exhaustive events are six { 1, 2, 3, 4, 5, 6 }
Equally Likely Event : Two or more events are said to be equally likely if the chances of their happening are equal.
eg. In throwing of an unbiased coin, result of Heat and Tail is equally likely.
Playing Cards:
(1) Total number of card are 52.
(2) There are 13 cards of each suit named Diamond, Hearts, Clubs and Spades
(3) Out of which Hearts and diamonds are red cards.
(4) Spades and Clubs are black cards
(5) There are four face cards each in number four Ace, King, Queen and Jack
Black Suit Card- (26)
ii) Club (13)
Red Suit Card–(26)
i) Diamond  (13)
ii) Heart (13)
(6)  Each Spade, Club, Diamond, Heart has 9 digit cards 2, 3, 4, 5, 6, 7, 8, 9 and 10
(7) There are 4 Honors cards each Spade, Club, Diamond, Heart contains 4 numbers of Honours cards Ace, King, Queen and Jack

Practice Set On Probability

1. A bag contains 6 red, 2 blue and 4 green balls. 3 balls are chosen at random. What is the probability that at least 2 balls chosen will be red?
A) 2/7
B) 1/2
C) 1/3
D) 2/5
E) 3/7

Option B
Solution:
There will be 2 cases
Case 1: 2 red, 1 blue orgreen
Prob. = 6C2 × 6C1 / 12C3 = 9/22
Case 2: all 3 red
Prob. = 6C3 / 12C3 = 2/22
Add the cases, required prob. = 9/22 + 2/22 = 11/22 = 1/2
2. Tickets numbered 1 to 250 are in a bag. What is the probability that the ticket drawn has a number which is a multiple of 4 or 7?
A) 83/250
B) 89/250
C) 77/250
D) 93/250
E) 103/250

Option B
Solution:
Multiples of 4 up to 120 = 250/4 = 62
Multiples of 7 up to 120 = 250/7 = 35 (take only whole number before the decimal part)
Multiple of 28 (4×7) up to 250 = 250/28 = 8
So total such numbers are = 62 + 35 – 8 = 89
So required probability = 89/250
3. From a deck of 52 cards, 3 cards are chosen at random. What is the probability that all are face cards?
A) 14/1105
B) 19/1105
C) 23/1105
D) 11/1105
E) 26/1105

Option D
Solution:
There are 3*4 = 12 face cards in 52 cards
So required probability = 12C3 / 52C3 = 11/1105
4. One 5 letter word is to be formed taking all letters – S, A, P, T and E. What is the probability that this the word formed will contain all vowels together?
A) 2/5
B) 3/10
C) 7/12
D) 3/5
E) 5/12

Option A
Solution:
Total words that can be formed is 5! = 120
Now vowels together:
Take: S, P, T and AE
So their arrangement is 4! * 2! = 48
So required probability = 48/120 = 2/5
5. One 5-digit number is to be formed from numbers – 0, 1, 3, 5, and 6 (repetition not allowed). What is the probability that number formed will be even?
A) 8/15
B) 7/16
C) 7/15
D) 3/10
E) 13/21

Option B
Solution:
Two cases:
Case 1: 0 at last place
So 4 choices for 1st digit, 3 for 2nd, 2 for 3rd and 1 for 4th. So numbers = 4*3*2*1 = 24
Case 2: 6 at last place
For 5-digit number 0 cannot be placed at 1st place or cannot be 1st digit
So 3 choices (1, 3, 5) for 1st digit, 3 for 2nd, 2 for 3rd and 1 for 4th. So numbers = 3*3*2*1 = 18
So total choices = 24+18 = 42
Number total 5-digit numbers that can be formed from 0, 1, 3, 5, and 6
0 not allowed at 1st place, so 4 choices for 1st place, 4 for 2nd, 3 for 3rd, 2 for 4th and 1 for 5th. Sp total = 4*4*3*2*1 = 96
So required probability = 42/96 = 7/16

Directions (6-8): There are 3 bags containing 3 colored balls – Red, Green and Yellow.
Bag 1 contains:
24 green balls. Red balls are 4 more than blue balls. Probability of selecting 1 red ball is 4/13

Bag 2 contains:
Total balls are 8 more than 7/13 of balls in bag 1. Probability of selecting 1 red ball is 1/3. The ratio of green balls to blue balls is 1 : 2

Bag 3 contains:
Red balls equal total number of green and blue balls in bag 2. Green balls equal total number of green and red balls in bag 2. Probability of selecting 1 blue ball is 3/14.

1. 1 ball each is chosen from bag 1 and bag 2, What is the probability that 1 is red and other blue?
A) 15/128
B) 21/115
C) 17/135
D) 25/117
E) 16/109

Option D
Solution:
Let red = x, so blue = x-4
So
x/(24+x+(x-4)) = 4/13
Solve, x = 16
So bag 1: red = 16, green  = 24, blue = 12
NEXT:
bag 2: total = 8 + 7/13 * 52 = 36
green and blue = y and 2y. Let red balls = z
So z + y + 2y = 36…………………(1)
Now Prob. of red = 1/3
So z/36 = 1/3
Solve, z = 12
From (1), y = 8
So bag 2: red = 12, green  = 8, blue = 16
NEXT:
bag 3: red = 8+16 = 24, green = 12+8 = 20
Blue prob. = 3/14
So a/(24+20+a) = 3/14
Solve, a = 12
So bag 3: red = 24, green  = 20, blue = 12
Now probability that 1 is red and other blue::
16/52 * 16/36 + 12/52 * 12/36 = 25/117
2. Some green balls are transferred from bag 1 to bag 3. Now probability of choosing a blue ball from bag 3 becomes 3/16. Find the number of remaining balls in bag 1.
A) 60
B) 58
C) 52
D) 48
E) 44

Option E
Solution:
blue balls in bag 3 are 12
Let x green balls are transferred. So
12/(56+x) = 3/16   [56 was number of bags in bag 3 before transfer] Solve, x = 8
So remaining number of balls in bag 1 = 52-8 = 44
3. Green balls in ratio 4 : 1 from bags 1 and 3 respectively are transferred to bag 4. Also 4 and 8 red balls from bags 1 and 3 respectively . Now probability of choosing green ball from bag 4 is 5/11. Find the number of green balls in bag 4?
A) 12
B) 15
C) 10
D) 9
E) 11

Option C
Solution:
4x and x = 5x green balls
4+8 = 12 red balls
So 5x/(5x+12) = 5/11
Solve, x = 2
5*2 = 10 green balls

Directions (9-10): There are 3 people – A, B and C. Probability that A speaks truth is 3/10, probability that B speaks truth is 3/7 and probability that C speaks truth is 5/6. For a particular question asked, at most 2 people speak truth. All people answer to a particular question asked.

1. What is the probability that B will speak truth for a particular question asked?
A) 7/18
B) 14/33
C) 4/15
D) 9/28
E) 10/33

Option D
Solution:
In any case B speaks truth. Now at most 2 people speak truth for 1 question
So case 1: B and A speaks truth
Probability = 3/7 * 3/10 * (1-5/6) = 3/140
Case 2: B and C speaks truth
Probability = 3/7 * ( 1- 3/10) * 5/6 = 5/20
Case 3: Only B speaks truth
Probability = 3/7 * ( 1- 3/10) * (1-5/6) = 1/20
Add the three cases = 6/20 + 3/140 = 45/140 = 9/28
2. A speaks truth only when B does not speak truth, then what is the probability that C does not speak truth on a question?
A) 11/140
B) 21/180
C) 22/170
D) 13/140
E) None of these

Option A
Solution:
Case 1: B does not speak truth, A speaks truth
So A speaks truth here
Probability that C does not speak truth = 3/10 * (1 – 3/7) * ( 1- 5/6) = 1/35
Case 2: B speaks truth
So A does not speak truth here
Probability that C does not speak truth = ( 1- 3/10) * 3/7 * ( 1- 5/6) = 1/20
So total = 1/35 + 1/20 = 11/140
• There are 100 tickets in a box numbered 1 to 100. 3 tickets are drawn at one by one. Find the probability that the sum of number on the tickets is odd.
A) 2/7
B) 1/2
C) 1/3
D) 2/5
E) 3/7

Option B
Solution:
There will be 4 cases
Case 1: even, even, odd
Prob. = 1/2 × 1/2 × 1/2
Case 2: even, odd, even
Prob. = 1/2 × 1/2 × 1/2
Case 3: odd, even, even
Prob. = 1/2 × 1/2 × 1/2
Case 4: odd, odd, odd
Prob. = 1/2 × 1/2 × 1/2
Add all the cases, required prob. = 1/2
• There are 4 green and 5 red balls in first bag. And 3 green and 5 red balls in second bag. One ball is drawn from each bag. What is the probability that one ball will be green and other red?
A) 85/216
B) 34/75
C) 95/216
D) 35/72
E) 13/36

Option D
Solution:
Case 1:first green, second red
Prob. = 4/9 × 5/8 = 20/72
Case 2:first red, second green
Prob. = 5/9 × 3/8 = 15/72
• A bag contains 2 red, 4 blue, 2 white and 4 black balls. 4 balls are drawn at random, find the probability that at least one ball is black.
A) 85/99
B) 81/93
C) 83/99
D) 82/93
E) 84/99

Option A
Solution:
Prob. (At least 1 black) = 1 – Prob. (None black)
So Prob. (At least 1 black) = 1 – (8C4/12C4) = 1 – 14/99
• Four persons are chosen at random from a group of 3 men, 3 women and 4 children. What is the probability that exactly 2 of them will be men?
A) 1/9
B) 3/10
C) 4/15
D) 1/10
E) 5/12

Option B
Solution:
2 men means other 2 woman and children
So prob. = 3C2 × 7C2 /10C4 = 3/10
• Tickets numbered 1 to 120 are in a bag. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?
A) 8/15
B) 5/16
C) 7/15
D) 3/10
E) 13/21

Option C
Solution:
Multiples of 3 up to 120 = 120/3 = 40
Multiples of 5 up to 120 = 120/5 = 24 (take only whole number before the decimal part)
Multiple of 15 (3×5) up to 120 = 120/15 = 8
So total such numbers are = 40 + 24 – 8 = 56
So required probability = 56/120 = 7/15
• There are 2 people who are going to take part in race. The probability that the first one will win is 2/7 and that of other winning is 3/5. What is the probability that one of them will win?
A) 14/35
B) 21/35
C) 17/35
D) 19/35
E) 16/35

Option D
Solution:
Prob. of 1st winning = 2/7, so not winning = 1 – 2/7 = 5/7
Prob. of 2nd winning = 3/5, so not winning = 1 – 3/5 = 2/5
So required prob. = 2/7 * 2/5 + 3/5 * 5/7 = 19/35
• Two cards are drawn at random from a pack of 52 cards. What is the probability that both the cards drawn are face card (Jack, Queen and King)?
A) 11/221
B) 14/121
C) 18/221
D) 15/121
E) 14/221

Option A
Solution:
There are 52 cards, out of which there are 12 face cards.
So probability of 2 face cards = 12C2/52C2 = 11/221
• A committee of 5 people is to be formed from among 4 girls and 5 boys. What is the probability that the committee will have less number of boys than girls?
A) 7/12
B) 7/15
C) 6/13
D) 5/12
E) 7/13

Option D
Solution:
Case 1: 1 boy and 4 girls
Prob. = 5C1 × 4C4/9C5 = 5/146
Case 2: 2 boys and 3 girls
Prob. = 5C2 × 4C3/9C5 = 40/126
Add the two cases = 45/126 = 5/12
• A bucket contains 2 red balls, 4 blue balls, and 6 white balls. Two balls are drawn at random. What is the probability that they are not of same color?
A) 5/11
B) 14/33
C) 2/5
D) 6/11
E) 2/3

Option E
Solution:
Three cases
Case 1: one red, 1 blue
Prob = 2C1 × 4C1 / 12C2 = 4/33
Case 2: one red, 1 white
Prob = 2C1 × 6C1 / 12C2 = 2/11
Case 3: one white, 1 blue
Prob = 6C1 × 4C1 / 12C2 = 4/11
• A bag contains 5 blue balls, 4 black balls and 3 red balls. Six balls are drawn at random. What is the probability that there are equal numbers of balls of each color?
A) 11/77
B) 21/77
C) 22/79
D) 13/57
E) 15/77

Option E
Solution:
5C2× 4C2× 3C212C6

Directions (1-3): An urn contains some balls colored white, blue and green. The probability of choosing a white ball is 4/15 and the probability of choosing a green ball is 2/5. There are 10 blue balls.

1. What is the probability of choosing one blue ball?
A) 2/7
B) 1/4
C) 1/3
D) 2/5
E) 3/7

Option C
Solution:
Probability of choosing one blue ball = 1 – (4/15 + 2/5) = 1/3
2. What is the total number of balls in the urn?
A) 45
B) 34
C) 40
D) 30
E) 42

Option D
Solution:
Probability of choosing one blue ball is 1/3
And total blue balls are 10. So with 10/30 we get probability as 1/3
So total balls must be 30
3. If the balls are numbered 1, 2, …. up to number of balls in the urn, what is the probability of choosing a ball containing a multiple of 2 or 3?
A) 3/4
B) 4/5
C) 1/4
D) 1/3
E) 2/3

Option E
Solution:
There are 30 balls in the urn.
Multiples of 2 up to 30 = 30/2 = 15
Multiples of 3 up to 30 = 30/3 = 10 (take only whole number before the decimal part)
Multiples of 6 (2×3) up to 30 = 30/6 = 5
So total such numbers are = 15 + 10 – 5 = 20
So required probability = 20/30 = 2/3
4. There are 2 brothers A and B. Probability that A will pass in exam is 3/5 and that B will pass in exam is 5/8. What will be the probability that only one will pass in the exam?
A) 12/43
B) 19/40
C) 14/33
D) 21/40
E) 9/20

Option B
Solution:
Only one will pass means the other will fail
Probability that A will pass in exam is 3/5. So Probability that A will fail in exam is 1 – 3/5 = 2/5
Probability that B will pass in exam is 5/8. So Probability that B will fail in exam is 1 – 5/8 = 3/8
So required probability = P(A will pass)*P(B will fail) + P(B will pass)*P(A will fail)
So probability that only one will pass in the exam = 3/5 * 3/8 + 5/8 * 2/5 = 19/40
5. If three dices are thrown simultaneously, what is the probability of having a same number on all dices?
A) 1/36
B) 5/36
C) 23/216
D) 1/108
E) 17/216

Option A
Solution:
Total events will be 6*6*6 = 216
Favorable events for having same number is {1,1,1}, {2,2,2}, {3,3,3}, {4,4,4}, {5,5,5}, {6,6,6} – so 6 events
Probability of same number on all dices is 6/216 = 1/36
6. There are 150 tickets in a box numbered 1 to 150. What is the probability of choosing a ticket which has a number a multiple of 3 or 7?
A) 52/125
B) 53/150
C) 17/50
D) 37/150
E) 32/75

Option E
Solution:
Multiples of 3 up to 150 = 150/3 = 50
Multiples of 7 up to 150 = 150/7 = 21 (take only whole number before the decimal part)
Multiples of 21 (3×7) up to 150 = 150/21 = 7
So total such numbers are = 50 + 21 – 7 = 64
So required probability = 64/150 = 32/75
7. There are 55 tickets in a box numbered 1 to 55. What is the probability of choosing a ticket which has a prime number on it?
A) 3/55
B) 5/58
C) 8/21
D) 16/55
E) 4/13

Option D
Solution:
Prime numbers up to 55 is 16 numbers which are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 43, 41, 47, 53.
So probability = 16/55
8. A bag contains 4 white and 5 blue balls. Another bag contains 5 white and 7 blue balls. What is the probability of choosing two balls such that one is white and the other is blue?
A) 61/110
B) 59/108
C) 45/134
D) 53/108
E) 57/110

Option D
Solution:
Case 1: Ball from first bag is white, from another is blue
So probability = 4/9 * 7/12 = 28/108
Case 1: Ball from first bag is blue, from another is white
So probability = 5/9 * 5/12 = 25/108
So required probability = 28/108 + 25/108 = 53/108
9. The odds against an event are 2 : 3 and the odds in favor of another independent event are 3 : 4. Find the probability that at least one of the two events will occur.
A) 11/35
B) 27/35
C) 13/35
D) 22/35
E) 18/35

Option B
Solution:
Let 2 events A and B
Odds against A are 2 : 3
So probability of occurrence of A = 3/(2+3) = 3/5. And non-occurrence of A = 2/5
Odds in favor of B are 3 : 4
So probability of occurrence of B = 3/(3+4) = 3/7. And non-occurrence of B = 4/7
Probability that at least one occurs
Case 1: A occurs and B does not occur
So probability = 3/5 * 4/7 = 12/35
Case 2: B occurs and A does not occur
So probability = 3/7 * 2/5 = 6/35
Case 3: Both A and B occur
So probability = 3/5 * 3/7 = 9/35
So probability that at least 1 will occur = 12/35 + 6/35 + 9/35 = 27/35
10. The odds against an event are 1 : 3 and the odds in favor of another independent event are 2 : 5. Find the probability that one of the event will occur.
A) 17/28
B) 5/14
C) 11/25
D) 9/14
E) 19/28

Option A
Solution:
Let 2 events A and B
Odds against A are 1 : 3
So probability of occurrence of A = 3/(1+3) = 3/4. And non-occurrence of A = 1/4
Odds in favor of B are 2 : 5
So probability of occurrence of B = 2/(2+5) = 2/7. And non-occurrence of B = 5/7
Case 1: A occurs and B does not occur
So probability = 3/4 * 5/7 = 15/28
Case 2: B occurs and A does not occur
So probability = 2/7 * 1/4 = 2/28
So probability that one will occur = 15/28 + 2/28 = 17/28
1. From a pack of 52 cards, 1 card is chosen at random. What is the probability of the card being diamond or queen?
A) 2/7
B) 6/15
C) 4/13
D) 1/8
E) 17/52

Option C
Solution:

In 52 cards, there are 13 diamond cards and 4 queens.
1 card is chosen at random
For 1 diamond card, probability = 13/52
For 1 queen, probability = 4/52
For cards which are both diamond and queen, probability = 1/52
So required probability = 13/52 + 4/52 – 1/52 = 16/52 = 4/13
2. From a pack of 52 cards, 1 card is drawn at random. What is the probability of the card being red or ace?
A) 5/18
B) 7/13
C) 15/26
D) 9/13
E) 17/26

Option B
Solution:

In 52 cards, there are 26 red cards and 4 ace and there 2 such cards which are both red and ace.
1 card is chosen at random
For 1 red card, probability = 26/52
For 1 ace, probability = 4/52
For cards which are both red and ace, probability = 2/52
So required probability = 26/52 + 4/52 – 2/52 = 28/52 = 7/13
3. There are 250 tickets in an urn numbered 1 to 250. One ticket is chosen at random. What is the probability of it being a number containing a multiple of 3 or 8?
A) 52/125
B) 53/250
C) 67/125
D) 101/250
E) 13/25

Option A
Solution:

Multiples of 3 up to 250 = 250/3 = 83 (take only whole number before the decimal part)
Multiples of 8 up to 250 = 250/3 = 31
Multiples of 24 (3×8) up to 250 = 250/24 = 10
So total such numbers are = 83 + 31 – 10 = 104
So required probability = 104/250 = 52/125
4. There are 4 white balls, 5 blue balls and 3 green balls in a box. 2 balls are chosen at random. What is the probability of both balls being non-blue?
A) 23/66
B) 5/18
C) 8/21
D) 7/22
E) 1/3

Option D
Solution:

Both balls being non-blue means both balls are either white or green
There are total 12 balls (4+3+5)
and total 7 white + green balls.
So required probability = 7C2/12C2 = [(7*6/2*1) / (12*11/2*1)] = 21/66 = 7/22
5. There are 4 white balls, 3 blue balls and 5 green balls in a box. 2 balls are chosen at random. What is the probability that first ball is green and second ball is white or green in color?
A) 1/3
B) 5/18
C) 1/2
D) 4/21
E) 11/18

Option B
Solution:

There are total 4+3+5 = 12 balls
Probability of first ball being green is = 5/12
Now total green balls in box = 5 – 1 = 4
So total white + green balls = 4 + 4 = 8
So probability of second ball being white or green is 8/12 = 2/3
So required probability = 5/12 * 2/3 = 5/18
6. 2 dices are thrown. What is the probability that there is a total of 7 on the dices?
A) 1/3
B) 2/7
C) 1/6
D) 5/36
E) 7/36

Option C
Solution:

There are 36 total events which can happen ({1,1), {1,2}……………….{6,6})
For a total of 7 on dices, we have – {1,6}, {6,1}, {2,5}, {5,2}, {3,4}, {4,3} – so 6 choices
So required probability = 6/36 = 1/6
7. 2 dices are thrown. What is the probability that sum of numbers on the two dices is a multiple of 5?
A) 5/6
B) 5/36
C) 1/9
D) 1/6
E) 7/36

Option E
Solution:

There are 36 total events which can happen ({1,1), {1,2}……………….{6,6})
For sum of number to be a multiple of 5, we have – {1,4}, {4,1}, {2,3}, {3,2}, {4,6}, {6,4}, {5,5} – so 7 choices
So required probability = 7/36
8. There are 25 tickets in a box numbered 1 to 25. 2 tickets are drawn at random. What is the probability of the first ticket being a multiple of 5 and second ticket being a multiple of 3.
A) 5/11
B) 1/4
C) 2/11
D) 1/8
E) 3/14

Option D
Solution:

There are 5 tickets which contain a multiple of 5
So probability of 1st ticket containing multiple of 5 = 5/25 = 1/5
Now:
Case 1: If the ticket chosen contained 15
If there was a 15 on first draw, then there are 7 tickets in box which contain a multiple of 3 out of 24 tickets. (25/3 – 1 = 8 – 1 = 7) – because 15 is already out from the box
So probability = 7/24 (24 tickets remaining after 1st draw)
Case 2: If the ticket chosen contained other than 15 (5 or 10 or 20 or 25)
If 15 was not there on first draw, then there are 8 tickets in box which contain a multiple of 3 out of 24 tickets. (25/3 = 8) – because 15 is already out from the box
So probability = 8/24 (24 tickets remaining after 1st draw)
Add the cases for probability of multiple of 3 on second ticket, so prob. = 7/24 + 8/24 = 15/24 (added the cases because we want one of these cases to happen and not both)
So required probability = 1/5 * 15/24 = 1/8 (multiplied the cases because we want both to happen)
9. What is the probability of selecting a two digit number at random such that it is a multiple of 2 but not a multiple of 14?
A) 17/60
B) 11/27
C) 13/30
D) 31/60
E) 17/30

Option C
Solution:

There are 90 two digit numbers (10-99)
Multiple of 2 = 90/2 = 45
Multiple of 14 = 90/14 = 6
Since all multiples of 14 are also multiple of 2, so favorable events = 45 – 6 = 39
So required probability = 39/90 = 13/30
10. There are 2 urns. 1st urn contains 6 white and 6 blue balls. 2nd urn contains 5 white and 7 black balls. One ball is taken at random from first urn and put to second urn without noticing its color. Now a ball is chosen at random from 2nd urn. What is the probability of the second ball being a white colored ball?
A) 11/13
B) 6/13
C) 5/13
D) 5/12
E) 11/12

Option A
Solution:

Case 1: first was a white ball
Now it is put in second urn, so total white balls in second urn = 5+1 = 6, and total balls in second urn = 12+1 = 13
So probability of white ball from second urn = 6/13
Case 2: first was a blue ball
Now it is put in second urn, so total white balls in second urn remain 5, and total balls in second urn = 12+1 = 13
So probability of white ball from second urn = 5/13
So required probability = 6/13 + 5/13 = 11/13 (added the cases because we want one of these cases to happen and not both)

1. A bag contains 12 white and 18 black balls. Two balls are drawn in succession without replacement. What is the probability that first is white and second is black?
A) 36/135
B) 36/145
C) 18/ 91
D) 30/91
E) None of these
2. Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?
A) 3/16
B) 1/8
C) 3/4
D) 1/2
E) None of these
3. In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected is:
A) 21/46
B) 21/135
C) 42/135
D) Can’t be determined
E) None of these
4. A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is?
A) 3/26
B) 3/52
C) 1/26
D) 1/4
E) None of these
5. A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are blue, is:
A)  1/91
B)  2/91
C) 3/91
D) 4/91
E) None of these.
6.  A bag contains 2 yellow, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?
A)  5/7
B) 1/21
C) 10/21
D) 2/9
E) None of these
7. Three coins are tossed. What is the probability of getting at most two tails?
A)  1/8
B) 5/8
C) 3/8
D) 7/8
E) None of these
8. One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card (Jack, Queen and King only)?
A)  1/13
B) 2/13
C) 3/13
D)  3/52
E) None of these
9. P and Q sit in a ring arrangement with 10 persons. What is the probability that P and Q will sit together?
A) 2/11
B) 3//11
C) 4/11
D) 5/11
E) None of these
10. Two dice are thrown simultaneously. Find the probability of getting a multiple of 2 on one dice and multiple of 3 on the other dice.
A)  1/9
B) 11/36
C) 13/36
E) None of these
1. B
2. C
3. A
4. C
5. D
6. C
7. D
8. C
9. A
10.BExplanation:1. The probability that first ball is white= 12c1/30c1= 2/5
Since, the ball is not replaced; hence the number of balls left in bag is 29.
Hence the probability the second ball is black= 18c1/29c1 =18/29
Required probability = 2/5*18/29 = 36/145

2. In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.
Then, E = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4),
(3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1),
(6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(E) = 27.
so probability = 27/36 = 3/4

3. Probability = 10c1*15c2/25c3
= 21/46

4. 2/52 = 1/26

5. 6c3/15c3 =4/91

6. 5c2/7c2 = 10/21

7. 7/8

8. 12/52 =3/13

9. n(S)= number of ways of sitting 12 persons at round table:
=(12-1)!=11!
Since two persons will be always together, then number of persons:
=10+1=11
So, 11 persons will be seated in (11-1)!=10! ways at round table and 2 particular persons will be seated in 2! ways.
n(A)= The number of ways in which two persons always sit together =10!×2
So probability = 10!*2!/11!= 2/11

10. 11/36

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