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 !!

**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**C) 33/221**

Explanation:

Case 1: 1 is king

^{4}C_{1}*^{48}C_{1}/^{52}C_{2}

Case 2: both are king

^{4}C_{2}/^{52}C_{2}

Add both cases.**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**B) 12/35**

Explanation:

Total balls = 15

Not blue means green or red i.e. any of (5+4) = 9 balls

So prob. =^{9}C_{2}/^{15}C_{2}**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**A) 4/9**

Explanation:

Case 1: Both blue

^{5}C_{2}/^{10}C_{2}

Case 2: Both white

^{5}C_{2}/^{10}C_{2}

Add both cases.**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**D) 10/21**

Explanation:

2 children should be there and rest 2 either from (2 women + 3 men) 5 people

So prob. =^{4}C_{2}*^{5}C_{2}/^{9}C_{4}**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**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**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**A) 55/221**

Explanation:

Prob. of both red =^{26}C_{2}/^{52}C_{2}

Prob. of both kings =^{4}C_{2}/^{52}C_{2}

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 =^{2}C_{2}/^{52}C_{2}

So required prob. =^{26}C_{2}/^{52}C_{2}+^{4}C_{2}/^{52}C_{2}–^{2}C_{2}/^{52}C_{2}**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**B) 16/45**

Explanation:

^{2}C_{1}*^{8}C_{1}/^{10}C_{2}**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**C) 49/225**

Explanation:

7/15 * 7/15**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**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.**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**B) 45/56**

Explanation:

2 men and 1 woman, prob. =^{5}C_{2}*^{3}C_{1}/^{8}C_{3}

2 women and 1 man, prob. =^{5}C_{1}*^{3}C_{2}/^{8}C_{3}

Add both cases

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