Hello Aspirants. Welcome to Online Quantitative Aptitude Section 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 bank exams !!

Questions Penned by Yogit

**A bag contains 5 red balls and 7 blue balls. Two balls are drawn at random without replacement, and then find the probability of that one is red and other is blue.**

a) 33/65

b) 35/66

c) 37/66

d) 41/65

e) None of theseAnswer –**b) 35/66**

**Explanation :**

(First red ball is drawn and then blue ball is drawn) + (first blue ball is drawn and then red ball is drawn)

(5/12)*(7/11) + (7/12)*(5/11) = 35/66**A bag contains 3 red balls and 8 blacks ball and another bag contains 5 red balls and 7 blacks balls, one ball is drawn at random from either of the bag, find the probability that the ball is red.**

a) 93/264

b) 95/264

c) 91/264

d) 97/264

e) None of theseAnswer –**c) 91/264**

**Explanation :**

Probability = probability of selecting the bag and probability of selecting red ball

(1/2)*(3/11) + (1/2)*(5/12) = 91/264**12 persons are seated at a circular table. Find the probability that 3 particular persons always seated together.**

a) 9/55

b) 7/55

c) 4/55

d) 3/55

e) None of theseAnswer –**d) 3/55**

**Explanation :**

total probability = (12-1)! = 11!

Desired probability = (10 – 1)! = 9!

So, p = (9! *3!) /11! = 3/55**P and Q are two friends standing in a circular arrangement with 10 more people. Find the probability that exactly 3 persons are seated between P and Q.**

a) 5/11

b) 4/11

c) 2/11

d) 3/11

e) None of theseAnswer –**c) 2/11**

**Explanation :**

Fix P at one point then number of places where B can be seated is 11.

Now, exactly three persons can be seated between P and Q, so only two places where Q can be seated. So, p = 2/11**A basket contains 5 black and 8 yellow balls. Four balls are drawn at random and not replaced. What is the probability that they are of different colours alternatively.**

a) 56/429

b) 57/429

c) 61/429

d) 68/429

e) None of theseAnswer –**a) 56/429**

**Explanation :**

sol=> BYBY + YBYB = (5/13)*(8/12)*(4/11)*(7/10) + (8/13)*(5/12)*(7/11)*(4/10) = 56/429**Direction(Q6 – Q8):**

**A bag contains 6 red balls and 8 green balls. Two balls are drawn at random one after one with replacement.****What is the probability that-**

**Both the balls are green**

a) 13/49

b) 15/49

c) 16/49

d) 17/49

e) None of theseAnswer –**c) 16/49**

**Explanation :**

P = (8/14)*(8/14)**First one is green and second one is red**

a) 16/49

b) 14/49

c) 11/49

d) 12/49

e) None of theseAnswer –**d) 12/49**

**Explanation :**

P = (8/14)*(6/14)**Both the balls are red**

a) 14/49

b) 9/49

c) 11/49

d) 12/49

e) None of theseAnswer –**b) 9/49**

**Explanation :**

P = (6/14)*(6/14)**Find the probability that in a leap year, the numbers of Mondays are 53?**

a) 1/7

b) 2/7

c) 3/7

d) 4/7

e) None of theseAnswer –**b) 2/7**

**Explanation :**

In a leap year there are 52 complete weeks i.e. 364 days and 2 more days. These 2 days can be SM, MT, TW, WT, TF, FS, and SS.

So P = 2/7**A urn contains 4 red balls, 5 green balls and 6 white balls, if one ball is drawn at random, find the probability that it is neither red nor white.**

a) 1/3

b) 1/4

c) 1/5

d) 2/3

e) None of theseAnswer –**a) 1/3**

**Explanation :**

5c1/15c1 = 1/3

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