Aptitude Questions: Probability Set 6

Hello Aspirants. Welcome to Online Maths in AffairsCloud.com. Here we are creating question sample in Probability, which is common for all the competitive exams. We have included Some questions that are repeatedly asked in exams !!

1. From a pack of 52 cads, 2 cards are drawn at random. What is the probability of drawing such that there is at least 1 king?
   A) 35/256
   B) 33/220
   C) 33/221
   D) 23/190
   E) 11/221
   
   Answer & Explanation

   C) 33/221
   Explanation:
   Case 1: 1 is king
   ⁴C₁*⁴⁸C₁ / ⁵²C₂
   Case 2: both are king
   ⁴C₂ / ⁵²C₂
   Add both cases.
   
   
   
2. A box contains 6 blue, 5 green and 4 red balls. If two balls are pick at random, then what is the probability that neither is blue?
   A) 10/21
   B) 12/35
   C) 3/5
   D) 11/21
   E) None of these
   
   Answer & Explanation

   B) 12/35
   Explanation:
   Total balls = 15
   Not blue means green or red i.e. any of (5+4) = 9 balls
   So prob. = ⁹C₂ / ¹⁵C₂
   
   
   
3. A box contains 5 blue and 5 white balls. What is the probability of drawing 2 balls such that both are same in color?
   A) 4/9
   B) 4/7
   C) 1/5
   D) 7/12
   E) 3/10
   
   Answer & Explanation

   A) 4/9
   Explanation:
   Case 1: Both blue
   ⁵C₂ / ¹⁰C₂
   Case 2: Both white
   ⁵C₂ / ¹⁰C₂
   Add both cases.
   
   
   
4. A committee of 4 people is to be formed from 3 men, 2 women and 4 children. What is the probability that exactly two of chosen people are children?
   A) 13/21
   B) 10/31
   C) 5/21
   D) 10/21
   E) None of these
   
   Answer & Explanation

   D) 10/21
   Explanation:
   2 children should be there and rest 2 either from (2 women + 3 men) 5 people
   So prob. = ⁴C₂*⁵C₂/ ⁹C₄
   
   
   
5. In a class 30% of the students opt for Math, 20% opt for Computers and 10% opt for both. A student is selected at random, find the probability that he has opted either Math or Computers.
   A) 3/5
   B) 2/5
   C) 4/9
   D) 6/11
   E) None of these
   
   Answer & Explanation

   B) 2/5
   Explanation:
   Prob. of math = 30/100 = 3/10, Prob. of computers = 20/100 = 1/5, prob. for both = 10/100 = 1/10
   So required prob. = 3/5 + 1/5 – 1/10
   
   
   
6. From a pack of 52 cards, 2 cards are drawn at random. What is the probability that either both are red or both are kings?
   A) 55/221
   B) 52/225
   C) 44/221
   D) 48/221
   E) None of these
   
   Answer & Explanation

   A) 55/221
   Explanation:
   Prob. of both red = ²⁶C₂/⁵²C₂
   Prob. of both kings = ⁴C₂/⁵²C₂
   Since there are also cads which are both red and king, so we will subtract there prob. There are 2 red cards which are kings
   Prob. of both red and king = ²C₂/⁵²C₂
   So required prob. = ²⁶C₂/⁵²C₂ + ⁴C₂/⁵²C₂ – ²C₂/⁵²C₂
   
   
   
7. A box contains 10 electric bulbs from which 2 bulbs are defective. Two bulbs are chosen at random. What is the probability that one of them is defective?
   A) 3/10
   B) 16/45
   C) 25/68
   D) 8/33
   E) 9/19
   
   Answer & Explanation

   B) 16/45
   Explanation:
   ²C₁*⁸C₁/¹⁰C₂
   
   
   
8. A bag contains 8 blue and 7 green balls. A ball is drawn out of it and put back in the bag. Then a ball is drawn again. What is the probability that both the balls are green?
   A) 36/225
   B) 48/221
   C) 49/225
   D) 40/221
   E) None of these
   
   Answer & Explanation

   C) 49/225
   Explanation:
   7/15 * 7/15
   
   
9. A bag contains 8 blue and 7 green balls. 2 balls are drawn one by one without replacement. What is the probability that the balls are alternately of different colors?
   A) 9/21
   B) 8/17
   C) 5/14
   D) 8/15
   E) 9/19
   
   Answer & Explanation

   D) 8/15
   Explanation:
   When 1st is blue, prob. = 8/15 * 7/14
   When 1st is green, prob. = 7/15 * 8/14
   Add both cases.
   
   
   
10. There are 5 men and 3 women. A committee of 3 members is to be made. Find the probability that either there are 2 men and 1 woman or 2 women and 1 man.
    A) 5/51
    B) 45/56
    C) 35/51
    D) 48/61
    E) None of these
    
    Answer & Explanation

    B) 45/56
    Explanation:
    2 men and 1 woman, prob. = ⁵C₂*³C₁/⁸C₃
    2 women and 1 man, prob. = ⁵C₁*³C₂/⁸C₃
    Add both cases