Quantitative Aptitude
Probability
1,860 Questions
Probability Questions
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$\dfrac {1}{8}$
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$\dfrac {1}{4}$
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$\dfrac {3}{8}$
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$\dfrac {1}{2}$
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$\dfrac {27}{32}$
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$\dfrac {5}{64}$
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$\dfrac {5}{32}$
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$\dfrac {1}{2}$
C
Correct answer
Explanation
This is a negative binomial distribution problem. We want the 4th blue ball on the 7th draw. This means in the first 6 draws, there are 3 blue balls and 3 green balls. The probability is (6C3) * (1/2)^3 * (1/2)^3 * (1/2) = 20 * (1/64) * (1/2) = 20/128 = 5/32.
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$\dfrac{4}{13}$
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$\dfrac{3}{13}$
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$\dfrac{2}{13}$
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none
A
Correct answer
Explanation
There are 13 spades and 4 kings in a deck. One card, the king of spades, is counted in both groups, so the total number of unique favorable cards is 13 + 4 - 1 = 16. The probability is 16/52, which simplifies to 4/13.
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(i)$\dfrac{1}{4}$ (ii) $\dfrac{2}{3}$
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(i)$\dfrac{4}{1}$ (ii) $\dfrac{2}{11}$
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(i)$\dfrac{5}{6}$ (ii) $\dfrac{5}{6}$
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(i)$\dfrac{13}{6}$ (ii) $\dfrac{2}{9}$
A
Correct answer
Explanation
Total balls = 4+5+3 = 12. (i) Probability of Red = 3/12 = 1/4. (ii) Probability of not Green = 1 - P(Green) = 1 - 4/12 = 1 - 1/3 = 2/3.
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$\dfrac {225}{18442}$
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$\dfrac {116}{20003}$
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$\dfrac {125}{15552}$
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None of these
C
Correct answer
Explanation
Probability of getting a 7 with two dice is 6/36 = 1/6. Probability of not getting a 7 is 5/6. Using binomial distribution for 4 successes in 6 trials: 6C4 * (1/6)^4 * (5/6)^2 = 15 * (1/1296) * (25/36) = 375 / 46656 = 125 / 15552.
B
Correct answer
Explanation
Total outcomes for two coins: {HH, HT, TH, TT}. Outcomes with one head and one tail: {HT, TH}. Probability = 2/4 = 1/2.
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$\dfrac {1}{3}$
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$\dfrac {1}{5}$
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$\dfrac {1}{10}$
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$\dfrac {1}{2}$
C
Correct answer
Explanation
A regular hexagon has 6 vertices. The total number of ways to choose 3 vertices is 6C3 = 20. There are only 2 sets of vertices that form an equilateral triangle (the two sets of alternating vertices). Thus, the probability is 2/20 = 1/10.
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$\displaystyle\frac{3}{27}$
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$\displaystyle\frac{3}{29}$
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$\displaystyle\frac{3}{25}$
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$\displaystyle\frac{3}{22}$
C
Correct answer
Explanation
Numbers from 2 to 101 inclusive. Total count = 101 - 2 + 1 = 100. Numbers less than 14 are {2, 3, ..., 13}. Count = 13 - 2 + 1 = 12. Probability = 12/100 = 3/25.
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$\displaystyle\frac{12}{2^5}$
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$\displaystyle\frac{13}{2^5}$
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$\displaystyle\frac{14}{2^5}$
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$\displaystyle\frac{15}{2^5}$
B
Correct answer
Explanation
The total number of outcomes for 5 coin tosses is 2^5 = 32. Using recursion or counting, the number of sequences without 'HH' is the (n+2)-th Fibonacci number, which for n=5 is F(7) = 13.
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$\dfrac {3}{4}$
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$\dfrac {1}{4}$
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$\dfrac {5}{8}$
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$\dfrac {2}{7}$
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$\dfrac{1}{256}$
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$\dfrac{7}{256}$
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$\dfrac{81}{256}$
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$\dfrac{31}{256}$
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$\dfrac{3}{8}$
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$\dfrac{2}{5}$
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$\dfrac{1}{10}$
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$\dfrac{4}{5}$
A
Correct answer
Explanation
Total outcomes for 4 coins = 2^4 = 16. Outcomes with 2 heads and 2 tails = 4C2 = 6. Probability = 6/16 = 3/8.
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$\dfrac{3}{8}$
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$\dfrac{49}{80}$
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$\dfrac{8}{13}$
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$\dfrac{1}{2}$
B
Correct answer
Explanation
P(W) = P(Bag1)P(W|Bag1) + P(Bag2)P(W|Bag2) = (1/2)(3/5) + (1/2)(5/8) = 3/10 + 5/16 = (24+25)/80 = 49/80.
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$\dfrac {55}{221}$.
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0
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Cant be determined
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None of the above
A
Correct answer
Explanation
Total cards = 52. Red cards = 26. Kings = 4. Red kings = 2. P(Red or King) = P(Red) + P(King) - P(Red and King) = 26/52 + 4/52 - 2/52 = 28/52 = 7/13. The question asks for two cards drawn. P(both red) = (26/52)(25/51) = 25/102. P(both kings) = (4/52)(3/51) = 1/221. P(both red and both kings) = P(both red kings) = (2/52)*(1/51) = 1/1326. Using inclusion-exclusion for two cards: 25/102 + 1/221 - 1/1326 = (325 + 6 - 1) / 1326 = 330/1326 = 55/221.