Probability Questions and Answers updated daily – Aptitude


Probability Questions and Answers updated daily – Aptitude



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

Probability Questions

1. 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 A's selection = 1/5
=> probability of A's Rejection = 1 - (1/5 ) = 4/5

And 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


2. 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)
=> 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


3. 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, 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)


4. In a class of 200 students, there are 80 boys and remaining 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


5. 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


6. A box contain 8 red balls, 6 green balls and 8 blue balls. A ball is drawn at random. What is the probability that the ball drawn is neither red nor green? 



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Correct Ans:(4 / 11)
Explanation:
Given
Total no. of balls =8+6+8 = 22 = n(S)
Event of getting a ball which is neither red nor green = a ball taken from 8 blue balls = 8C1 = 8 = n(E)

RequiredProbability (neither red nor green) = n(E) / n(S) = 8/22 = 4/11


7. 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) = (5,6) and (6,5)
=> n(E) = 2


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


8. A dice is thrown twice . What is the probability that at least one of them comes up with number 4 



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Correct Ans:(11 / 36)
Explanation:
When two dice are thrown the possible cases are (1,1), (1,2), ... (6,6) there are 36 such possible outcomes.
For 4 to comes atleast once (4,1) , (4,2) , (4,3) , (4,4) , (4,5) , (4,6) , (1,4) , (2,4) , (3,4) , (5,4) , (6,4)
There are 11 such cases.
P(E) =n(E) /n(S)
where as F(s) = no of favourable chances
T(s) = Total no of cases.
Probability = 11 / 36.


9. 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:
5 digit number = 5! = 120
Divisible by 5 then the last digit should be 0
Then the remaining position have the possibility = 4! =24
P = (4!/5!) = 24/120 = 1/5.


10. 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


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



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Correct Ans:1/3
Explanation:
MATRIX : 6 words, 2 vowel and 4 consonants
Probability of choosing a vowel = 2 / 6 = 1/3


12. 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(S1)=52
after drawing one black card,total sample space =n(S2)=51
event of drawing 2 black cards successively = n(E1)=26C1 = 26
n(E2)= 25C1=25

Hence probability = [n(E1)/n(S1)] X[n(E2)/n(S2)]
= (26/52) x (25/51)
= 25/102
Thus required probability of drawing 2 black cards successively=25/102


13. 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.


14. A bag contains 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white?  



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Correct Ans:(4/7)
Explanation:
Let number of balls = (6 + 8) = 14.
Number of white balls = 8.
Probability (drawing a white ball) = 8/14 = 4/7


15. 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


16. 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 .


17. In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?  



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Correct Ans:(2/7)
Explanation:
Given,In a lottery, there are 10 prizes and 25 blanks.
So, Total number of sample space = 10 + 25 = 35
=> n(S) = 35

Event of getting prize = n(E) = 10

Required probability = P (getting a prize) = n(E) / n(S)
=10 / 35
= 2/7


18. What is the probability of getting a sum 9 from two throws of a dice? 



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Correct Ans:(1/9)
Explanation:
In two throws of a dice, n(S) = (6 x 6) = 36.
Let E = event of getting a sum "9" = {(3, 6), (4, 5), (5, 4), (6, 3)}.
Therefore, P(E)= n(E)/n(S) = 4/36 = 1 / 9


19. A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue? 



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Correct Ans:(10/21)
Explanation:
Total number of balls = (2 + 3 + 2) = 7.

Let S be the sample space.
Then, n(S) = Number of ways of drawing 2 balls out of 7 = 7c2
= ( 7 x 6 ) / (2x1)
= 7 x 3
= 21 .

Let E = Event of drawing 2 balls, none of which is blue.
Therefore, n (E) = Number of ways of drawing 2 balls out of (2 red + 3 green) balls.
= 5c2
= (5 x 4 )/(2 x 1)
= 5 x 2
= 10

Required Probability, P(E) =n(E) / n (s)
= 10/21


20. In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?



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Correct Ans:(1 / 3)
Explanation:
Total number of balls = (8 + 7 + 6) = 21. Let E = event that the ball drawn is neither red nor green = event that the ball drawn is blue. Therefore , n (E) = 7 Therefore P (E) = n (E) / n (S) = 7/21 = 1/3.



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