Welcome to Online Quantitative Aptitude Section in AffairsCloud.com. Here we are creating question sample in Probability, which is common for all competitive exams. We have included Some questions that are repeatedly asked in bank exams !!
- A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both hearts. Find the Probability of the lost card being a heart?
E. None of theseAnswer – C. 11/50
Total cards = 52
Drawn cards(Heart) = 2
Present total cards = total cards-drawn cards =52-2=50
Remaining Card 13-2 = 11
Probability = 11/50
- There are three boxes each containing 3 Pink and 5 Yellow balls and also there are 2 boxes each containing 4 Pink and 2 Yellow balls. A Yellow ball is selected at random. Find the probability that Yellow ball is from a box of the first group?
E. None of theseAnswer – B. 45/61
Probability = (3/5 * 5/8)/([3/5 * 5/8] + [2/5 * 1/3]) = 45/61
- A fruit basket contains 10 Guavas and 20 Bananas out of which 3 Guavas and 5 Bananas are defective. If two fruits selected at random, what is the probability that either both are Bananas or both are non-defective?
E. None of theseAnswer – D. 316/435
P(A) = 20c2 / 30c2, P(B) = 22c2 / 30c2
P(A∩B) = 15c2 / 30c2
P(A∪B) = P(A) + P(B) – P(A∩B) => (20c2/30c2)+(22c2/30c2)-(15c2/30c2)=316/435
- A committee of five persons is to be chosen from a group of 10 people. The probability that a certain married couple will either serve together or not at all is?
E. None of theseAnswer – D. 51/126
Five persons is to be chosen from a group of 10 people = 10C5 = 252
Couple Serve together = 8C3 * 2C2 = 56
Couple does not serve = 8C5 = 56
Probability = 102/252 = 51/126
- Out of 14 applicants for a job, there are 6 women and 8 men. It is desired to select 2 persons for the job. The probabilty that atleast one of selected persons will be a Woman is?
E. None of theseAnswer – A. 77/91
Man only = 8C2 = 14
Probability of selecting no woman = 14/91
Probability of selecting atleast one woman = 1 – 14/91 = 77/91
- Three Bananas and three oranges are kept in a box. If two fruits are chosen at random, Find the probability that one is Banana and another one is orange?
E. None of theseAnswer – B. 3/5
Total probability = 6C2 = 15
Probability that one is Banana and another one is orange = 3C1 * 3C1 = 9
probability = 9/15 = 3/5
- A basket contains 6 White 4 Black 2 Pink and 3 Green balls. If three balls picked up random, What is the probability that all three are White?
D. 8/92Answer – A. 4/91
Total Balls = 15
Probability = 6c3 / 15c3 = 4/91
- A basket contains 6 White 4 Black 2 Pink and 3 Green balls. If three balls are picked at random, what is the probability that two are Black and one is Green?
E. 18/455Answer – E. 18/455
Total Balls = 15
Probability = 4c2 * 3c1/ 15c3 = 18/455
- A basket contains 6 White 4 Black 2 Pink and 3 Green balls. If four balls are picked at random, what is the probability that atleast one is Black?
E. None of theseAnswer – A. 69/91
Total Balls = 15
Probability = 11c4/15c4 = 22/91
One is black = 1 – 22/91 = 69/91
- A basket contains 6 White 4 Black 2 Pink and 3 Green balls.If two balls are picked at random, what is the probability that either both are Pink or both are Green?
E. None of theseAnswer – B. 4/105
Probability both are Pink = 1/15C2
Probability both are Green = 3/15C2
Required Probability = 4/15C2 = 4/105