# Probability Short Tricks & Tips

Probability Short-Cut Tricks & Tips : Probability Short-Cut Tricks & Tips Question Pdf for Banking, SSC, RRB, FCI, Railway, UPSC, State PCS, Insurance & other Competitive exams. Probability Short-Cut Tricks & Tips shortcut Tricks Pdf, Probability Short-Cut Tricks & Tips MCQ, Probability Short-Cut Tricks & Tips Objective Question & Answer Pdf. “Probability Short-Cut Tricks & Tips Questions PDF” In this post we are providing you the Probability Short-Cut Tricks & Tips pdf with detailed solution & Short Tricks. So that you can easily get the logic of question. This Probability Short-Cut Tricks & Tips Pdf we are Providing is free to download. ” Most Important Probability Short-Cut Tricks & Tips Question PDF with Answers

Probability Short-Cut Tricks & Tips Plays a vital role in Exam. In every exam you will get at least 5-10 questions from this topic. So candidates must focus on this topic and download this Probability Short-Cut Tricks & Tips pdf to get important questions with best solution regarding Probability Short-Cut Tricks & Tips. We have put all Previous Year Questions of Probability Short-Cut Tricks & Tips that are Asked in various Govt & Private Exam.

In everyday life, we come across the situations having either some certainty or uncertainty and we are generally interested to measure this certainty or uncertainty. We want to measure that up to what extent this particular situation would occur. We generally achieve it qualitatively; not able to calculate it quantitatively. So what about if we want to measure it quantitatively? That is achieved with the help of the Theory of Probability.

Probability is the Measure of Uncertainty.

Experiment:An operation which can produce some well-defined outcomes is called an experiment.

Random Experiment:

An experiment in which all possible outcomes are know and the exact output cannot be predicted in advance, is called a random experiment.

Examples:

1. Rolling an unbiased dice.
2. Tossing a fair coin.
3. Drawing a card from a pack of well-shuffled cards.
4. Picking up a ball of certain colour from a bag containing balls of different colours.
⇒When we throw a coin, then either a Head (H) or a Tail (T) appears.

⇒A dice is a solid cube, having 6 faces, marked 1, 2, 3, 4, 5, 6 respectively. When we throw a die, the outcome is the number that appears on its upper face.

⇒A pack of cards has 52 cards.

It has 13 cards of each suit, name Spades, Clubs, Hearts and Diamonds.

Cards of spades and clubs are black cards.

Cards of hearts and diamonds are red cards.

There are 4 honours of each unit.

There are Kings, Queens and Jacks. These are all called face cards.

Sample Space:

When we perform an experiment, then the set S of all possible outcomes is called the sample space

In tossing a coin, S = {H, T}

If two coins are tossed, the S = {HH, HT, TH, TT}.

In rolling a dice, we have, S = {1, 2, 3, 4, 5, 6}.

Probability of Occurrence of an Event:
Let S be the sample space and E be the event
Then

1) ?(?) = ?
2) ≤ ??(?) ≤ ?
3) ?(?) = ? → ? ?? ?????? ? ???? ?????
4) ?(?) = ? → ? ?? ?????? ??????????
5) ?(?) + ?(?̅) = 1

EXAMPLE:-A fair coin is tossed at random. Find the probability of getting:
2) Tail

EXAMPLE:- Two unbiased coins are tossed simultaneously at random.
Find the probability of getting:
1) Head on the first coin.
2) Head on the second coin.
3) Head on both the coins.

1. Lets take an example..
Set your timer for 60 seconds, and see if you can get the correct answer. Then read below for the full explanation, and see if you’ve fallen into a trap or not!

Question – Sixty percent of the members of a study group are women, and 45 percent of those women are lawyers. If one member of the study group is to be selected at random, what is the probability that the member selected is a woman lawyer?

(A) 0.10
(B) 0.15
(C) 0.27
(D) 0.33
(E) 0.45

Although the question uses the word “probability,” the concept it tests for and the trap laid within are percent-related—the piece of information you need from the wide field of probability to solve this problem is the basic formula:

probability = number of wanted outcomes / total number of outcomes

In essence, probability, like a percentage, is a ratio between a part and a whole, expressed as a fraction.

So why did this relatively simple problem catch my eye? Precisely because it is deceptively easy, which is why a decent percent of GMAT test takers will get it wrong, at least when put under a time crunch. For many test takers, the following (mistaken) thought pattern will ensue:

60% of the members are women. Imagine a pie chart, with a 60% chunk marked “women.”

Now, 45% are lawyers. Take a chunk of 45% (almost half of the pie), out of the original 60% chunk, and that’s your percent of women lawyers—45%, or 0.45 (answer choice E).

In extreme rush cases, a test taker may even forget what he’s looking for in the first place. Once you’re imagining a 45% chunk taken out of the 60% chunk, it’s deceptively easy to fall into the trap of focusing on what remains: a 15% “slice,” which will lead the test-taker in a hurry into choosing B in the rush to move on to the next question.

Both of these thought processes and the resulting answer choices are wrong. The stumbling point that the GMAT test-writers are counting on is the failure to ask the simple questions whenever the word “percent” appears: What is the whole? What number or quantity is the percent taken out of?

The first percent (60% women) is indeed taken out of the members of the study group. The next line has a crucial phrase: 45% of those women are lawyers. So the next percent is not taken out of the entire pie chart, but out of the 65% chunk alone. We’re looking for 45% of the group titled women, which happens to be given as a percent of the whole, not just 45% of the entire group.

The actual calculation is therefore 45% of 60%, or {45/100}*{60/100} (think of any “of” in these cases as a multiplication sign).

One last note: instead of actually calculating the above expression, just ballpark it. The group you seek (women lawyers) is ‘slightly less than half’ of the women (as 45% is just under 50%). The right answer will therefore need to be something slightly smaller than {1/2}*0.6 = 0.3. Only one answer choice out of the five answer choices presented fits that description, and that is C 0.27. Answer choices A and B are too small, and D and E are already over half of 60%.

2. Consider the following question:
A canoe has two oars, left and right. Each oar either works or breaks. The failure or non-failure of each oar is independent of the failure or non-failure of the other. You can still row the canoe with one oar. The probability that the left oar works is 3/5. The probability that the right oar works is also 3/5. What is the probability that you can still row the canoe?
A) 9/25
B) 10/25
C) 6/10
D) 2/3
E) 21/25

1. The wrong way
The temptation is to multiply the two probabilities given to reach the answer 9/25. Whenever you get to an answer choice very quickly, particularly when that answer is A, I would look at the question again! Answer choice A is the first answer you see. If you are in a hurry and option A looks right, many test takers will go for A.

This calculation only gives you the probability that both oars work.
To get the right answer, you would also have to add the probability that the left oar works and the right fails.
Then you would also have to add the probability that the right works, but the left fails.
All this would be possible, but slow. Is there a better way? Yes!

2. The right way
Simply look at the question from the other side. What is the probability that you can’t row the canoe? This would be {2/5}*{2/5} = 4/25.
Using the idea that the probability of something happening is 1- the probability that it doesn’t happen, you can use the following equation to reach the right answer: 1 – 4/25 = 21/25. Answer choice E.
We call this the Forbidden Method: subtracting the ‘forbidden’ or unwanted probabilities from the total probability, which is 1.

When two dice are thrown The probability of sample points for the given sum can be known with below logic.

Total No. of sample points when two dice are thrown  = (6*6) =36

Question 1: Find the probability to get sum 3 when two dice are thrown simultaneously?

Sol :      P(S)  =  P(3) / P(Total) = 2/36 = 1/13.

Note: ( From the figure we know for sum 3 the probability of sample points are 2)

Question 2: Find the probability to get sum 10  when two dice are thrown simultaneously?

Sol: Probability to get sum 10 = P(10)/P(total )

= 3/36 ( From the figure we know for sum 10 the probability of sample points are 3)

When three dice are thrown The probability of sample points for the given sum can be known with below logic.

Example Question: Find the probability to get sum 5  when three dice are thrown simultaneously?

Sol :

Probability to get sum 5 = P(5)/P(total no. of sample points )

= 6/216.

Note: ( From the above figure we know for sum 5 the probability of sample points are 6)

# 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 = 36/1452. 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 = 3/43. Probability = 10c1*15c2/25c3
= 21/464. 2/52 = 1/265. 6c3/15c3 =4/916. 5c2/7c2 = 10/217. 7/88. 12/52 =3/139. 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!= 2/1110. 11/36