Probability Questions and Answers updated daily – Aptitude

Probability Questions: Solved 93 Probability Questions and answers section with explanation for various online exam preparation, various interviews, Aptitude Category online test. Category Questions section with detailed description, explanation will help you to master the topic.

Probability Questions

21. A military man can strike a target once in 3 bullets. If he fires 3 bullets in succession, what is the probability that he will strike his target?




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Correct Ans:(19/27)
Explanation:
Probability of hitting the target is the sum of the probability of hitting in (case i) any 1 of the 3 bullets ; (case ii) any 2 of the 3 bullets or (case iii) all of the 3 bullets.

So, the only case where the man will not strike the target is when he fails to strike the target even in one of the three bullets that he takes.

The probability that he will not strike the target in one bullet = 1 - Probability that he will strike target in exact one bullet (out of 3 bullets)
= 1 - (1/3)
= 2/3

Therefore, the probability that he will not strike the target in all the three bullets
= (2/3) * (2/3) * (2/3)
= 8/27

Hence, the probability that he will strike the target at least in one of the three bullets = 1 - (8/27)
= 19/27
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22. Three dice are thrown together. Find the probability of getting a total of at least 6 ?




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Correct Ans:103/108
Explanation:
Since one die can be thrown in six ways to obtain any one of the six numbers marked on its six faces
⇒ Total number of elementary events = 6 x 6 x 6 = 216
Let A be the event of getting a total of at least 6.Then Ā denotes the event of getting a total of less than 6 i.e. 3, 4, 5.
⇒ Ā = { (1,1,1), (1,1,2), (1,2,1), (2,1,1), (1,1,3),(1,3,1), (3,1,1), (1,2,2), (2,1,2), (2,2,1) }
So, favorable number of cases = 10
⇒ P(Ā) = 10/216
1 – P (A) = 10/216
P(A) = 1 – 10/216 = 103/108
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23. A number is selected at random from first 40 natural numbers. What is the chance that it is a multiple of either 4 or 14? 




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Correct Ans:11/40
Explanation:
We know that,
Probability=Favorable Cases/Total Cases
The probability that the number is a multiple of 4 is 10/40
Since favorable cases here {4,8,12,16,20,24,28,32,36,40} = 10 cases
Total cases= 40 cases
Similarly the probability that the number is a multiple of 14 is 2/40.
Since favorable cases here {14,28} =2 cases
Total cases= 40 cases
Multiple of 4 or 14 have common multiple from 1 to 40 is 28.
Hence, these events are mutually exclusive events.
Therefore chance that the selected number is a multiple of 4 or 14 is:
= (10+2-1)/40 = 11/40
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24. In a bag there are 4 white, 4 red and 2 green balls. Two balls are drawn at random. What is the probability that at least one ball is of green colour?




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Correct Ans:2/5
Explanation:
Total number of balls in the bag = 4 + 4 + 2 = 10
Total possible outcomes = Selection of 2 balls out of 10 balls
= 10C2 = (10*9)/(1*2) = 45
Favourable outcomes = 1 green ball and 1 ball of other colour + 2 green balls
= 2C1 * 8C1 + 2C2
= 2 * 8 + 2 = 18
Therefore, required probability
= 18/45 = 2/5
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25. An urn contains 9 blue, 5 white and 7 black balls. If 2 balls are drawn at random, what is the probability that only one ball is white ?




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Correct Ans:8/21
Explanation:
Total balls in the urn = 9 + 5 + 7 = 21
One ball will be white while other ball will be either blue or black.
Total possible outcomes = Selection of 2 balls out of 21 balls = 21C2 = (21*20)/(1*2) = 210
Total favourable outcomes = Selection of one ball out of 5 white balls and selection of 1 ball out of 16 remaining balls.
= 5C1 * 16C1 = 5*16 = 80
Required probability = 80/210 = 8/21
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26. There are 7 red balls and 8 yellow balls in a bag. Two balls are simultaneously drawn at random. What is the probability that both the balls are of same colour ?




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Correct Ans:7/15
Explanation:
Total balls in the bag = 7 + 8 = 15
Total possible outcomes = Selection of 2 balls out of 15 balls
= 15C2 = (15*14) / (1*2) = 105
Total favourable outcomes = Selection of 2 balls out of 8 yellow balls + Selection of 2 balls out of 7 red balls
= 8C2 + 7C2 = [(8*7)/(1*2)] + [(7*6)/(1*2)]
= 28 + 21 = 49
Required probability = 49/105 = 7/15
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27. A boss decides to distribute Rs. 2000 between 2 employees. He knows X deserves more that Y, but does not know how much more. So he decides to arbitrarily break Rs. 2000 into two parts and give X the bigger part. What is the chance that X gets twice as much as Y or more? 




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Correct Ans:2/3
Explanation:
Given X receives the bigger part. The bigger part could be any number from 1000 to 2000.

Now, if the bigger part is to be at least twice as much as the smaller part, we have

X ≥ 2Y
---> X ≥ 2(2000 – X)
---> X ≥ 4000 - 2X
---> 3X ≥ 4000
---> X ≥ 4000/3

Now, Probability that X lies between 4000/3 and 2000 = (2000−4000/3)/(2000−1000)
= [(6000 - 4000)/3]/1000
= [2000]/3000
= 2/3
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28. A book-shelf contains 2 English, 3 Hindi and 4 Sanskrit books. If two books are picked at random, then what is the probability that either all are Sanskrit or all are English books? 




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Correct Ans:7/36
Explanation:
Total number of books = 2 + 3 + 4 =9
Probability of choosing 2 books = 9Câ‚‚
= (9 x 8)/(1 x 2)
= 36
Probability of choosing either Sanskrit or English books = 4Câ‚‚ + 2Câ‚‚
= [(4 x 3)/2] + 1
= 6 + 1
= 7
Required probability = 7/36.
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29. In a college, 35% of the students speak English, 55% speak Hindi and 14(2/7)% speak both the languages. If a student is selected at random for a competition, what is the probability that he/she can speak English or Hindi?




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Correct Ans:53/70
Explanation:
Let total number of students be 100%.
Probability of students who speak English P(E) = 35/100 = 7/20
Probability of students who speak Hindi P(H) = 55/100 = 11/20
Probability of students who speak both Hindi and English P(E â‹‚ H) = (14(2/7))/100 = 1/7
Probability of student who can speak English or Hindi = P(E) + P(H) - P(E â‹‚ H)
= (7/20) + (11/20) - (1/7)
= 18/20 - 1/7
= 9/10 - 1/7
= (63 - 10)/70
= 53/70
Hence, the required probability is 53/70.
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30. In a class test of 50 students, 20 are boys and 30 are girls. 10 boys and 15 girls ranked in A grade. If a student is chosen at random from the class, what is the probability of choosing a girl or an A grade student? 




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Correct Ans:4/5
Explanation:
Given:
Total number of students = 50
Number of boys = 20
Number of girls = 30
Probability of choosing a girl = 30C₁/50C₁
= 30/50 = 3/5

Also given,
Number of A grade boys = 10
Number of A grade girls = 15
Probability of choosing A grade student = 25C₁/50C₁
= 25/50 = 1/2

Probability of choosing grade A girl = 15C₁/50C₁
= 15/50 = 3/10

Therefore, probability of choosing a girl or A grade student = [(3/5) + (1/2)] - (3/10)
= [(6+5)/10] - (3/10)
= (11/10) - (3/10)
= 8/10 = 4/5.
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31. A basket contains 4 roses, 3 lilies and 2 sunflowers. A man picks 2 flowers at random from the basket. What is the probability that both are roses or both are lilies?




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Correct Ans:1/4
Explanation:
The total number of ways of selecting 2 flowers out of 9 flowers
= 9Câ‚‚ = (9 x 8)/(1 x 2)
= 36
The total number of ways of selecting 2 roses = 4Câ‚‚
= (4 x 3)/(1 x 2)
= 6

The total number of ways of selecting 2 lilies = 3Câ‚‚
= (3 x 2)/(2 x 1)
= 3
Required probability of both are roses or both are lilies = (4Câ‚‚/ 9Câ‚‚) + (3Câ‚‚/9Câ‚‚)
=(6/36) + (3/36)
= 9/36
= 1/4.
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32. There are 12 blue pens and 18 black pens. Two pens one after another are taken out without replacement. What is the probability that the first pen is of blue colour and the second one is of black colour?




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Correct Ans:36/145
Explanation:
Total number of pens = 12 + 18 = 30

Required Probability = (12/30) x (18/29)
= 36/145
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33. The probability of drawing a red card from a deck of playing cards is ?




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Correct Ans:1/2
Explanation:
Total number of cards, n(S) = 52
Number of red cards, n(E) = 13 + 13 = 26
P(E) = n(E)/n(S)
= 26/52
= 1/2
Therefore, the probability of drawing a red card P(E) = 1/2
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34. Three dice are thrown together. Find the probability of getting a total of at least 6 ?




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Correct Ans:103/108
Explanation:
Since one dice can be thrown in six ways to obtain any one of the six numbers marked on its six faces.
⇒ Total number of elementary events = 6 x 6 x 6 = 216
Let A be the event of getting a total of at least 6.Then Ā denotes the event of getting a total of less than 6 i.e. 3, 4, 5.
⇒ Ā = { (1,1,1), (1,1,2), (1,2,1), (2,1,1), (1,1,3),(1,3,1), (3,1,1), (1,2,2), (2,1,2), (2,2,1) }
So, favorable number of cases = 10
⇒ P(Ā) = 10/216
⇒ 1 – P (A) = 10/216
⇒ P(A) = 1 – (10/216)
= 103/108
Hence, option (C) is correct.
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35. A coin is tossed thrice. The probability that exactly two heads show up is 




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Correct Ans:3/8
Explanation:
A possible outcomes of tossing a coin thrice are
{HHH, HHT, HTH, THH, HTT, TTH, THT, TTT}
Therefore, total number of outcomes n(S) = 8
Number of favourable outcomes = {HHT, HTH, THH}
Therefore, total favourable outcomes n(P) = 3
Hence, probability of getting exactly two heads = n(P)/n(S) = 3/8
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36. A basket contains 5 white and 9 black balls. There is another basket which contains 7 white and 7 black balls. One ball is to drawn from either of the two baskets. What is the probability of drawing a black ball?




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Correct Ans:4/7
Explanation:

Probability of choosing 1 basket = ½

Probability of black ball from 1st basket = (1/2)*(9C1/14C1)
= (1/2)*(9/14) = 9/28

Probability of black ball from 2nd basket = (1/2)*(7C1/14C1)
= (1/2)*(7/14) = 7/28

Required probability = (9/28) + (7/28) = 16/28 = 4/7.

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37. Two dice are thrown. Find the probability of getting a sum of number which is prime?




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Correct Ans:5/12
Explanation:
Prime numbers will be 2, 3, 5, 7 and 11.
Probability of getting prime numbers from two dice,
2 - (1,1)
3 - (1,2), (2,1)
5 - (1,4), (2,3), (3,2), (4,1)
7 - (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
11 - (5,6), (6,5)
Total possible ways = 15
Total probability = 15/36 = 5/12
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38. A box contains 7 black and 5 white hair clips. 3 hair clips are drawn at random. What is the probability that one is black and the other 2 are white?




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Correct Ans:7/22
Explanation:
Let S be the sample space. Then,
n(S) = no.of ways of drawing 3 hair clips out of 12 = 12C₃
= (12*11*10)/(3*2*1)
= 220
Let E be event of drawing 1 black and 2 white hair clips.
n(E) = no. of ways of drawing 1 black of 7 and 2 white out of 5.
n(E) = 7C₁ * 5C₂
= 7*[(5*4)/(1*2)]
= 70
Required probability = 70/220 = 7/22
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39. In a box there are 10 apples and 2/5th of the apples are rotten. If three apples are taken out from the box, what will be the probability that at least one apple is rotten.




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Correct Ans:5/6
Explanation:
Total number of apples = 10
Number rotten apples = 10*(2/5) = 4
So, number of good apples = 10 - 4 = 6
If 1 apple is rotten and 2 apples are good,
= 4C1 * 6C2
= (4/1) * [(6*5)/(1*2)]
= 60
If 2 apple is rotten and 1 apples are good,
= 4C2 * 6C1
= [(4*3)/(1*2)] * (6/1)
= 36
If 3 apples are rotten,
= 4C3
= [(4*3*2)/(1*2*3)]
= 4
Total outcomes = 10C3
= [(10*9*8)/(1*2*3)]
= 120
Probability = (60 + 36 + 4)/120
= 100/120
= â…š
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40. If the letters of the word SUPPOSE are placed in a row then what is the probability that the vowels always come together?




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Correct Ans:1/7
Explanation:
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