CPT Section D, Quantitative Aptitude, Chapter 13 Prof(Dr.) P.R.Vittal

(a)2/7 b)3 /7 c)4/7 (d)None of these Solution :A leap year contains 52 weeks plus 2 days.The least 2 days can be [s,m],m,tu],[tu,wed],[wed,thurs],[thurs,fri],[fri,sat],[sat,sun] P(53 sun or mon )=3/7 Answer: (b)

(a)5:7 (b)4:7 (c)5:8 (d)4:5 Solution : P(A)=7/12 No. of favourable ways =7 , No. of unfavourable ways =5 Hence odds against the event A is 5:7 Answer: (a)

(a)1/15 (b)1/12 (c)5/18 (d)5/21 Solution : The probability that all the 3 are white =(5C3/10C3)=1/12 Answer : (b)

(a)5/396 (b)3/132 (c)1/36 (d)1/22 Solution : P(3 are red) = ((3C3).(9C2)/12C5)=1/22 Answer : (d)

(a)0.6 (b)0.06 (c)0.006 (d)0.96 Solution : P(A∩B∩C)=P(A)P(B)P(C)=0.06 Answer : (b)

There are 10 balls numbered from 1 to 10 in a box. If one of them is selected at random , the probability that the number printed on the ball would be an even number greater than 5 is (a)3/5 (b)2/5 (c)3/10 (d)7/10 Solution : The set of even numbers >5 is (6,8,10). p(ball has even number>5)=3/10 Ans : (c)

For two independent events A and B , P(A)=1/5 and P(AUB)=3/4. Then P(B) is (a)11/20 (b)11/16 (c)5 /15 (d)9/20 Solution : P(AUB)=P(A)+P(B)-P(A)P(B) ¾ = 1/5 +P(B)-(1/5)P(B) P(B)=11/16 Ans : (b)

A problem on probability is given to two students A and B. Their chances of solving it are 1/3 and ¼.The probability that the problem is solved is (a)11/12 (b)1/12 (c)3/4 (d)1/2 Solution : P(A)=1/3, P(A’)=2/3 , P(B)=1/4 , P(B’)=3/4 P(problem solved)=1-P(A’)P(B’)=1-(2/3)(¾) =1/2 Answer : (d)

If X is a random variable such that X

: 1

P(x) : 0

2

3

4

k/10 3k/10 3k/5

The value of k is (a)1 (b)2 (c)10

(d)5

Solution : Since ∑p(x)=1 , 0+k/10 + 3k/10 + 3k/5 =1 K=1 Answer: (a)

The random variable X has the following distribution X P(x)

:

0 :

1 0

2k

2

3

3k

k

Then p(x