# 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

41. Two urns contain 5 white and 7 black balls and 3 white and 9 black balls respectively. One ball is transferred to the second urn and then one ball is drawn from the second urn. Find the probability that the first ball transferred is black, given that the ball drawn is black?

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Correct Ans:14/23

Explanation:

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42. Four unit squares are chosen at random on a chessboard. What is the probability that three of them are of one colour and fourth is of opposite colour?

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Correct Ans:640/1281

Explanation:

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43. In a certain game, you perform three tasks. You flip a coin and success would be head. You roll a single dice and success would be a six. You pick a card from a full playing-card deck, and success would be picking a spade card. If any of these task are successful, then you win the game. What is the probability of winning?

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Correct Ans:11/16

Explanation:

In this scenario, winning combinations would include success on any one task as well as any combination of two or three successes.In other words, there are several cases that constitute the winning combinations.By contrast, the only way to lose the game would be unsuccessful at all three tasks.Let’s use the**complement rule**.

P(lose game) = P(Coin = T AND dice ≠ 6 AND card ≠ spades)

= (1/2)*(5/6)*(3/4) = 5/16

P(win game) = 1 – P(lose game) = 1 – (5/16) = 11/16

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44. There are 8 boys and 12 girls in a class. 5 students have to be chosen for an educational trip. Find the number of ways in which this can be done if 2 particular girls are always included.

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Correct Ans:816

Explanation:

18c3 = 816 (2 girls already selected).

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45. A box contains either blue or red flags. The total number of flags in the box is an even number. A group of children are asked to pick up two flags each. If all the flags are used up in the process such that 60% of the children have blue flags, and 55% have red flags, what percentage of children have flags of both the colors?

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Correct Ans:15%

Explanation:

In this problem, let A be the event that a child picks up a blue flag, B be the event that a child picks up a red flag. Then, A∪B is the event that the child picks up either a blue or a red flag, and A∩B is the event that the child picks up both a blue and a red flag.

From the problem statement we know:

I. P(AUB) = 100% (Since there are even number of flags and all the flags in the box are taken by the children)

II. P(A) = 60%

III.P(B) = 55%

IV. P(A∩B)=?

We are trying to find the percentage of children who have flags of both the colors. Substituting the values from (3.) into the equation given in (1.) we get:

P(A∪B) = P(A) + P(B) - P(A∩B)

100% = 60% + 55% - P(A∩B)

P(A∩B) = (60% +55%)-100%

P(A∩B) =115%-100%

P(A∩B) =15%

Therefore the percentage of children who got both a red flag and a blue flag is 15%. Thus, C is the correct answer choice.

From the problem statement we know:

I. P(AUB) = 100% (Since there are even number of flags and all the flags in the box are taken by the children)

II. P(A) = 60%

III.P(B) = 55%

IV. P(A∩B)=?

We are trying to find the percentage of children who have flags of both the colors. Substituting the values from (3.) into the equation given in (1.) we get:

P(A∪B) = P(A) + P(B) - P(A∩B)

100% = 60% + 55% - P(A∩B)

P(A∩B) = (60% +55%)-100%

P(A∩B) =115%-100%

P(A∩B) =15%

Therefore the percentage of children who got both a red flag and a blue flag is 15%. Thus, C is the correct answer choice.

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46. In a certain game, you pick a card from a standard deck of 52 cards. If the card is a heart, you win. If the card is not a heart, the person replaces the card to the deck, reshuffles, and draws again. The person keeps repeating that process until he picks a heart, and the point is to measure: how many draws did it take before the person picked a heart and won? What is the probability that one will have at least three draws before one picks a heart?

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Correct Ans:9/16

Explanation:

A full deck of 52 cards contains 13 cards from each of the four suits. The probability of drawing a heart from a full deck is 1/4. Therefore, the probability of “not heart” is 3/4.

P(at least three draws to win) = 1 – P(win in two or fewer draws)

Furthermore,

P(win in two or fewer draws) = P(win in one draw OR win in two draws)

= P(win in one draw) + P(win in two draws)

Winning in one draw means: I select one card from a full deck, and it turns out to be a heart. Above, we already said: the probability of this is 1/4.

P(win in one draw) = 1/4

Winning in two draws means: my first draw is “not heart”, P = 3/4, AND the second draw is a heart, P = 1/4. Because we replace and re-shuffle, the draws are independent, so the AND means multiply.

P(win in two draws) =(3/4)*(1/4) = 3/16

P(win in two or fewer draws) =P(win in one draw) + P(win in two draws)

= 1/4 + 3/16 = 7/16

P(at least three draws to win) = 1 – P(win in two or fewer draws)

= 1 – 7/16 = 9/16.

P(at least three draws to win) = 1 – P(win in two or fewer draws)

Furthermore,

P(win in two or fewer draws) = P(win in one draw OR win in two draws)

= P(win in one draw) + P(win in two draws)

Winning in one draw means: I select one card from a full deck, and it turns out to be a heart. Above, we already said: the probability of this is 1/4.

P(win in one draw) = 1/4

Winning in two draws means: my first draw is “not heart”, P = 3/4, AND the second draw is a heart, P = 1/4. Because we replace and re-shuffle, the draws are independent, so the AND means multiply.

P(win in two draws) =(3/4)*(1/4) = 3/16

P(win in two or fewer draws) =P(win in one draw) + P(win in two draws)

= 1/4 + 3/16 = 7/16

P(at least three draws to win) = 1 – P(win in two or fewer draws)

= 1 – 7/16 = 9/16.

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47. There are 450 junior students and 300 senior students. Among these students, there are 40 sibling pairs where each pair has 1 junior and 1 senior. One student is chosen from senior and one from junior randomly. What is the probability that the two students selected are from a sibling pair?

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Correct Ans:316/13500

Explanation:

Given,

No. of Junior students = 450

No. of Senior students = 300

There are 40 sibling pair, each containing 1 junior and 1 senior

No ofstudents in40 sibling pair = 2 x 40 = 80 students

One student is chosen from senior

= 300C

= 300

One student is chosen from junior

=450C

= 450

Therefore, sample space for one student chosen from senior and one student chosen from junior:-

n(s) = 300 x 450 = 135000

Event of selecting two students from a sibling pair

n(E) = 80C

= (80 * 79) / 2

= 3160

Probability that the two selected students are from a sibling pair

P(E) = n(E) / n(S)

= 3160 / 135000

=

No. of Junior students = 450

No. of Senior students = 300

There are 40 sibling pair, each containing 1 junior and 1 senior

No ofstudents in40 sibling pair = 2 x 40 = 80 students

One student is chosen from senior

= 300C

_{1}= 300

One student is chosen from junior

=450C

_{1}= 450

Therefore, sample space for one student chosen from senior and one student chosen from junior:-

n(s) = 300 x 450 = 135000

Event of selecting two students from a sibling pair

n(E) = 80C

_{2}= (80 * 79) / 2

= 3160

Probability that the two selected students are from a sibling pair

P(E) = n(E) / n(S)

= 3160 / 135000

=

**316/13500**
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48. Two person A and B appear in an interview. The probability of A's selection is 1/5 and the probability of B's selection is 2/7. What is the probability that only one of them is selected?

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Correct Ans:(13 / 35)

Explanation:

Given,

Probability of

=> probability of

And the probability of

=> probability of

= A selects and B rejects (+) B selects and A rejects

= (1/5)*(5/7) + (2/7)*(4/5)

= (1/7) + (8/35)

= (5 + 8) / 35

=

Thus,

Probability of

**A's selection**= 1/5=> probability of

**A's Rejection**= 1 - (1/5 ) = 4/5And the probability of

**B's selection**= 2/7=> probability of

**B's Rejection**= 1 - (2/7) = 5/7**Probability that only one of them is selected = A selects and B rejects (OR) B selects and A rejects**= A selects and B rejects (+) B selects and A rejects

= (1/5)*(5/7) + (2/7)*(4/5)

= (1/7) + (8/35)

= (5 + 8) / 35

=

**13 / 35**Thus,

**the Probability that only one of them is selected = 13 / 35**
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49. There are 12 boys and 8 girls in a tuition centre. If three of them scored first mark, then what is the probability that one of the three is a girl and the other two are boys?

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Correct Ans:( 44 / 95 )

Explanation:

Given

Total number of students = 20

Let S be the sample space.

Then, n(S) = number of ways of three scored first mark

n(S) = 20C3

= ( 20 * 19 * 18 ) / ( 3 * 2 * 1)

= (20 * 19 * 3)

=>

Let, E be the event of one is a girl and the other two are boys out of three students scored first mark.

Therefore,

n(E) = 8C1 * 12C2

= ( 8 ) * (12 * 11) / ( 2* 1)

= (8 * 6 * 11)

=>

Required probability,

= ( 8 * 6 * 11 ) / (20 * 19 * 3)

=

Total number of students = 20

Let S be the sample space.

Then, n(S) = number of ways of three scored first mark

n(S) = 20C3

= ( 20 * 19 * 18 ) / ( 3 * 2 * 1)

= (20 * 19 * 3)

=>

**n(S) = (20 * 19 * 3)**Let, E be the event of one is a girl and the other two are boys out of three students scored first mark.

Therefore,

**n(E) = number of possible of 1 girl out of 8 and 2 boys out of 12**n(E) = 8C1 * 12C2

= ( 8 ) * (12 * 11) / ( 2* 1)

= (8 * 6 * 11)

=>

**n(E) = (8 * 6 * 11)**Required probability,

**P(E) = n(E) / n(S)**= ( 8 * 6 * 11 ) / (20 * 19 * 3)

=

**44 / 95**
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50. A box contains 3 white, 4 red and 7 blue erasers. If five erasers are taken at random then the probability that all the five are blue color is:

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Correct Ans:( 3 / 286 )

Explanation:

Given

Total number of erasers in the box = 3 + 4 + 7 = 14.

Let S be the sample space.

Then, n(S) = number of ways of taking 5 erasers out of 14.

Therefore,

= 14 x 13 x 11

=

Let E be the event of getting all the 5 blue erasers out of 7 blue erasers.

Therefore,

=

Now, the required probability

= 21/2002

= 3/286

The

Total number of erasers in the box = 3 + 4 + 7 = 14.

Let S be the sample space.

Then, n(S) = number of ways of taking 5 erasers out of 14.

Therefore,

**n(S)**= 14C5 = (14 x 13 x 12 x 11 x 10) / (2 x 3 x 4 x 5)= 14 x 13 x 11

=

**2002**Let E be the event of getting all the 5 blue erasers out of 7 blue erasers.

Therefore,

**n(E)**= 7C5 = (7 x 6 x 5 x 4 x 3) / (2 x 3 x 4 x 5)=

**21**Now, the required probability

**P(E) = n(E) / n(S)**= 21/2002

= 3/286

The

**probability that all the five erasers are blue color = (3 / 286)**
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51. In a class of 200 students, there are 80 boys and remaining are girls. A student is selected at random, find the probability of selecting a girl from the students?

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Correct Ans:3 / 5

Explanation:

Out of 200 students, there are 80 boys and 120 girls.

Selecting a girl, the probability is = 120/200 = 3 / 5

Selecting a girl, the probability is = 120/200 = 3 / 5

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52. A bag contain 6 white balls and 4 red balls .Three balls are drawn randomly.What is the probability that one ball is red and other 2 are white?

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Correct Ans:(1 / 2)

Explanation:

Total number of balls = 6 + 4 = 10

3 balls drawn = 10C3 =(10 x 9 x 8)/(3 x 2 x 1) =120

one ball is red and other 2 are white = 4C1 x 6C2

= 4C1 x 6C2 = (4 x 6 x 5)/(1 x 2) = 60

Required Probability, P = 60/120 = 1 / 2

3 balls drawn = 10C3 =(10 x 9 x 8)/(3 x 2 x 1) =120

one ball is red and other 2 are white = 4C1 x 6C2

= 4C1 x 6C2 = (4 x 6 x 5)/(1 x 2) = 60

Required Probability, P = 60/120 = 1 / 2

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53. Two dice are thrown simultaneously, what is the probability of getting a sum of 11?

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Correct Ans:(1/18)

Explanation:

Total number of sample space (when two dice are thrown simultaneously) = n(S) = 36

Events on getting a sum of 11 (when two dice are thrown simultaneously) =

Thus,

=> Probability = 2/ 36 = 1/18

Events on getting a sum of 11 (when two dice are thrown simultaneously) =

**(5,6) and (6,5)**

=> n(E) = 2=> n(E) = 2

Thus,

**Required Probability P(E) = n(E) / n(S)**=> Probability = 2/ 36 = 1/18

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54. A five digit number is formed with the digits 0,1,2,3 and 4 without repetition. Find the chance that the number is divisible by 5.

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Correct Ans:(1 / 5)

Explanation:

Total numbers that can be formed using the given 5 digits without repetition = 5! = (5 x 4 x 3 x 2 x 1) =120

Number thus formed with the given digits 0,1,2,3 and 4, to be divisible by 5, the last digit should be 0.

Remaining 4 digits can be arranged in the first four positions in 4! ways = 24

Therefore Required Probability, P(E) = (4!/5!) = 24/120 = 1/5.

Number thus formed with the given digits 0,1,2,3 and 4, to be divisible by 5, the last digit should be 0.

Remaining 4 digits can be arranged in the first four positions in 4! ways = 24

Therefore Required Probability, P(E) = (4!/5!) = 24/120 = 1/5.

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55. A word is formed by taking all the letters from the word "MATRIX".In the new word formed, what is the probability that the vowels will be together ?

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Correct Ans:1/3

Explanation:

Given word : MATRIX

We can form 6! = 720 words.

For the vowels to be together, treat (AI) , M R T X => which means there are total 5 words, which can formed in 5! = 120 words

Multiply this by 2, since the vowels can be AI or IA, hence there are totally 240 words.

Probability = 240 / 720 = 1/3

We can form 6! = 720 words.

For the vowels to be together, treat (AI) , M R T X => which means there are total 5 words, which can formed in 5! = 120 words

Multiply this by 2, since the vowels can be AI or IA, hence there are totally 240 words.

Probability = 240 / 720 = 1/3

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56. A letter is chosen at random from the given word "MATRIX". What is the probability that the letter chosen is a vowel?

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Correct Ans:1/3

Explanation:

Total Number of letters in the given word "MATRIX" = 6

=> n(s) = 6

The given word "MATRIX" contains

Then the

= 2/6

=

=> n(s) = 6

The given word "MATRIX" contains

**2 vowels**(ie., A and I) and 4 consonants (ie., M, T, R, X)Then the

**Probability**that the letter chosen is a vowel = Number ofvowels in"MATRIX" /Total Number of letters in"MATRIX"= 2/6

=

**1/3**
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57. Find the probability of selecting two black cards drawn from a well shuffled pack of cards successively without replacement.

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Correct Ans:25/102

Explanation:

There are totally 52 cards and

and there are 26 black cards.

Given, two black cards are drawn successively,

total sample space =n(S

after drawing one black card,total sample space =n(S

event of drawing 2 black cards successively = n(E

n(E

Hence

= (26/52) x (25/51)

= 25/102

and there are 26 black cards.

Given, two black cards are drawn successively,

total sample space =n(S

_{1})=52after drawing one black card,total sample space =n(S

_{2})=51event of drawing 2 black cards successively = n(E

_{1})=26C1 = 26n(E

_{2})= 25C1=25Hence

**probability = [n(E**_{1})/n(S_{1})] X[n(E_{2})/n(S_{2})]= (26/52) x (25/51)

= 25/102

**Thus required probability of drawing 2 black cards successively=25/102**
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58. Find the probability of getting 53 sundays on a leap year?

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Correct Ans:2/7

Explanation:

Leap Year consists of 366 days = 52 Full Weeks + 2 days

This 2 day can be Sun-Mon, Mon-Tue, Tue-Wed, Wed-Thu, Thu-Fri, Fri-Sat or Sat-Sun

The probaility is 2/7.

This 2 day can be Sun-Mon, Mon-Tue, Tue-Wed, Wed-Thu, Thu-Fri, Fri-Sat or Sat-Sun

The probaility is 2/7.

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59. 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)?

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Correct Ans:(3/13)

Explanation:

Clearly, there are 52 cards, out of which there are 12 face cards.

**Probability of getting a face card**= 12/52 =**3/13**
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60. A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is:

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Correct Ans:(1/26)

Explanation:

Here, n(S) = 52.
Let E = event of getting a queen of club or a king of heart.
Then, n(E) = 2.
Therefore, P(E) = n (E)/n(S) = 2/52 = 1/26 .

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