Quantitative Aptitude
Probability
1,860 Questions
Probability Questions
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$0.50$
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$0.40$
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$0.60$
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$0.30$
D
Correct answer
Explanation
The balls are numbered 1 to 10. The odd numbers are 1, 3, 5, 7, 9. Those greater than 4 are 5, 7, and 9. There are 3 such balls out of 10, so the probability is 3/10 = 0.30.
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$0.1$
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$\dfrac{1}{4}$
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$0.3$
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None
B
Correct answer
Explanation
A standard deck has 52 cards, and there are 13 cards of each suit. The probability of drawing a spade is 13/52, which simplifies to 1/4.
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$\dfrac{1}{2}$
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$\dfrac{1}{3}$
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$\dfrac{1}{6}$
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$\dfrac{5}{6}$
A
Correct answer
Explanation
A die has 6 faces: {1, 2, 3, 4, 5, 6}. Even numbers are {2, 4, 6}. Probability = 3/6 = 1/2.
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0
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1
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$\frac{1}{3}$
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$\frac{2}{3}$
A
Correct answer
Explanation
A die has outcomes {1, 2, 3, 4, 5, 6}. Numbers less than 3 are {1, 2}. Numbers greater than 2 are {3, 4, 5, 6}. There is no number that is both less than 3 and greater than 2. Probability = 0.
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$\dfrac{1}{2}$
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$\dfrac{21}{50}$
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$\dfrac{29}{50}$
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$\dfrac{42}{58}$
C
Correct answer
Explanation
Experimental probability = (Number of favorable outcomes) / (Total number of trials). Head = 58, Total = 100. Probability = 58/100 = 29/50.
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$\dfrac{4}{15}$
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$\dfrac{1}{3}$
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$\dfrac{2}{5}$
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$\dfrac{1}{5}$
A
Correct answer
Explanation
Total tosses = 600. The number of times one head occurred is 160. The probability is 160 / 600. Simplifying the fraction: 16/60 = 4/15.
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$ \dfrac{29}{50}$
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$ \dfrac{21}{50}$
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$ \dfrac{11}{25}$
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$ \dfrac{14}{25}$
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$\dfrac{1}{6}$
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$\dfrac{2}{6}$
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$\dfrac{3}{6}$
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$\dfrac{4}{6}$
C
Correct answer
Explanation
A dice has 6 outcomes: 1, 2, 3, 4, 5, 6. Prime numbers are 2, 3, 5. There are 3 prime numbers. Probability = 3/6.
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a prime number
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a multiple of $3$ and $5$
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an odd number
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neither divisible by $5$ nor by $10$
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$\dfrac {1}{8}$
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$\dfrac {1}{6}$
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$\dfrac {2}{8}$
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$\dfrac {1}{5}$
A
Correct answer
Explanation
Each toss has 2 outcomes. For 3 tosses, total outcomes = 2^3 = 8. The only outcome with all heads is (H, H, H), which is 1 outcome. Probability = 1/8.
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$\dfrac{{^{39}{C_4}}}{{^{52}{C_4}}}$
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$\dfrac{{^{12}{C_3}}}{{^{52}{C_4}}}$
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$ - \dfrac{{^{39}{C_4}}}{{^{52}{C_4}}}$
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$\dfrac{{^{12}{C_4}}}{{^{52}{C_4}}}$
B
Correct answer
Explanation
Total ways to choose 4 cards is 52C4. We want 4 spades, but one must be a king. There is only 1 king of spades. So we must pick the king of spades (1C1) and 3 other spades from the remaining 12 spades (12C3). Probability = 12C3 / 52C4.
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$\dfrac{1}{99}$
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$\dfrac{1}{22}$
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$\dfrac{1}{34}$
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$\dfrac{3}{44}$
D
Correct answer
Explanation
Total balls = 3+4+5 = 12. Probability of drawing 3 white balls = (3/12)(2/11)(1/10) = 6/1320. Probability of 3 green = (4/12)(3/11)(2/10) = 24/1320. Probability of 3 red = (5/12)(4/11)(3/10) = 60/1320. Sum = (6+24+60)/1320 = 90/1320 = 9/132 = 3/44.
$3$ players A,B & C toss a coin cyclically in the order (that is A, B, C, A, B, C, A, B.....) till a head shows. Let $p$ be the probability that the coin shows a head. Let $\alpha ,\beta\ &\ \gamma $ be respectively the probabilities that A, B and C gets the first head. Determine $\alpha $ (in terms of p)
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$\displaystyle \frac{(1-p)p}{1-(1-p)^3}$
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$\displaystyle \frac{p}{1-(1-p)^3}$
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$\displaystyle \frac{(1-p)^2p}{1-(1-p)^3}$
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none
B
Correct answer
Explanation
Let p be the probability of head. A wins if head occurs on 1st, 4th, 7th... toss. Probability = p + (1-p)^3 * p + (1-p)^6 * p + ... This is a geometric series with first term p and common ratio (1-p)^3. Sum = p / (1 - (1-p)^3).
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$0.19$
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$0.39$
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$0.29$
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$0.25$
C
Correct answer
Explanation
The probability that a randomly selected valve is non-defective is 0.9 for factory A and 0.8 for factory B. Using the mixture of 4 valves from A and 5 from B, the probability that two selected valves are both non-defective is approximately 0.71. Thus the probability of at least one defective valve is approximately 1 - 0.71 = 0.29.
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$\displaystyle \frac{1}{6^{6}}$
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$\displaystyle \frac{ 6!}{6^{6}}$
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$\displaystyle \frac{ 6!}{ 3! \; 3!}\times \displaystyle \frac{1}{6^{6}}$
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$1$
C
Correct answer
Explanation
This is a multinomial distribution problem. Probability of 2 is 1/6, 4 is 1/6, other is 4/6. We want 3 of 2, 3 of 4, 0 of other in 6 trials. Formula: (6! / (3!3!0!)) * (1/6)^3 * (1/6)^3 * (4/6)^0 = (6! / (3!3!)) * (1/6)^6.