Quantitative Aptitude
Probability
1,860 Questions
Probability Questions
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$\displaystyle{\frac{2}{4}}$
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$\displaystyle{\frac{2}{6}}$
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$\displaystyle{\frac{2}{7}}$
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none of these
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$\displaystyle \frac{2}{3}$
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$\displaystyle \frac{1}{2}$
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$\displaystyle \frac{1}{6}$
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$\displaystyle \frac{5}{6}$
B
Correct answer
Explanation
The prime numbers on a standard die are 2, 3, and 5. There are 3 prime numbers out of 6 total outcomes. The probability is 3/6 = 1/2.
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$\displaystyle \frac{13}{15}$
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$\displaystyle \frac{11}{15}$
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$\displaystyle \frac{3}{5}$
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$\displaystyle \frac{2}{5}$
C
Correct answer
Explanation
Total probability must be 1. P(white) = 1 - (P(red) + P(blue)) = 1 - (2/15 + 4/15) = 1 - 6/15 = 9/15 = 3/5.
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$\dfrac 12$
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$\dfrac 34$
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$\dfrac 56$
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Can't be determined
A
Correct answer
Explanation
There are 20 tickets numbered 1 to 20. The odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19. There are 10 odd numbers. The probability is 10/20 = 1/2.
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$\displaystyle\frac{1}{2}$
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$\displaystyle\frac{10}{21}$
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$\displaystyle\frac{11}{21}$
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$\displaystyle\frac{13}{21}$
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$\displaystyle \frac{1}{26}$
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$\dfrac1{13}$
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$\dfrac{1}{52}$
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$\dfrac{2}{13}$
A
Correct answer
Explanation
There are 2 red Queens (Hearts and Diamonds) in a deck of 52 cards. Probability = 2 / 52 = 1 / 26.
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$\dfrac{1}{6}$
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$\dfrac{1}{3}$
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$\dfrac{2}{3}$
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$\dfrac56$
A
Correct answer
Explanation
A standard die has 6 faces. The probability of rolling a 3 is 1 favorable outcome out of 6 total outcomes.
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$\dfrac{1}{2}$
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$\dfrac13$
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$\dfrac23$
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$\dfrac14$
A
Correct answer
Explanation
A standard die has outcomes {1, 2, 3, 4, 5, 6}. Even numbers are {2, 4, 6}. Probability is 3/6 = 1/2.
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$\displaystyle \frac{1}{13}$
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$\displaystyle \frac{1}{52}$
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$\displaystyle \frac{1}{4}$
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none of these
A
Correct answer
Explanation
There are 13 spades in a deck. Only one of them is an ace. Given the card is a spade, the probability it is an ace is 1/13.
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non of these
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$1$
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$0$
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can not determined
C
Correct answer
Explanation
There are 20 balls in total. The probabilities of red, black, and green are 5/20, 7/20, and 8/20 respectively. Red therefore does not have the greatest probability, so the statement is false and the required entry is 0.
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$\displaystyle \frac{1}{2}$
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$\displaystyle \frac{1}{3}$
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$\displaystyle \frac{1}{4}$
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$\displaystyle \frac{5}{6}$
B
Correct answer
Explanation
Multiples of 3 as sums: 3, 6, 9, 12. Ways: 3 (2 ways), 6 (5 ways), 9 (4 ways), 12 (1 way). Total = 2+5+4+1 = 12. Probability = 12/36 = 1/3.
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$\dfrac{1}{3}$
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$\dfrac{1}{2}$
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$\dfrac{1}{4}$
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$\dfrac{1}{5}$
B
Correct answer
Explanation
A die has 6 faces: 1, 2, 3, 4, 5, 6. The odd numbers are 1, 3, 5. Probability = 3/6 = 1/2.
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$\dfrac{1}{5}$
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$\dfrac{2}{5}$
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$\dfrac{3}{5}$
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$\dfrac{4}{5}$
B
Correct answer
Explanation
The prime numbers between 1 and 20 are 2, 3, 5, 7, 11, 13, 17, 19. There are 8 such numbers. The probability is 8/20 = 2/5.
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$\displaystyle\frac{1}{36}$
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$\displaystyle\frac{5}{36}$
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$\displaystyle\frac{11}{36}$
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$\displaystyle\frac{25}{36}$
D
Correct answer
Explanation
Probability of 5 appearing on one die is 1/6. Probability of not appearing is 5/6. For two dice, probability of 5 not appearing on either is (5/6) * (5/6) = 25/36.
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$\displaystyle \frac {1}{2}, \frac {3}{4}$
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$\displaystyle \frac {2}{3}, \frac {1}{4}$
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$\displaystyle \frac {1}{4}, \frac {4}{5}$
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$\displaystyle \frac {1}{2}, \frac {2}{3}$
A
Correct answer
Explanation
For two coin tosses, the sample space is {HH, HT, TH, TT}. Exactly one head occurs in {HT, TH}, so the probability is 2/4 = 1/2. At least one head occurs in {HH, HT, TH}, so the probability is 3/4.