Multiple Choice Questions

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 2 3 4 P(x) : 0 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 : 0 1 2 3 P(x) : 0 2k 3k k Then p(x<2) is (a)1/6 (b)1/3 (c)2/3 (d)0 Solution : Since ∑p(x)=1 , 0+2k+3k+ k=1 k=1/6 Hence p(x<2)=0+2k=2/6=1/3 Ans : (b)

If X is a continuous random variable with pdf

f(x)= c/x2 , 2<x<∞

0 , otherwise The value of c is

(a)1 (b)2 (c)4 (d)3

Solution :

2∞∫f(x)dx=1 hence ∫c/x^2 dx =1

C=2

Answer : (b)

The odds in favour of two mutually exclusive events are 2:5 and 3:7. The probability that atleast one of two events to occur is

(a)3/35 (b)1/2 (c)41/70 (d)29/70

Solution :

P(A)=2/7 p(B)=3/10p(AuB)=2/7+3/10 =41/70

Ans : (c)

The probability that a person has a B.Com degree is 0.88 and he is CA is 0.30. The probability that he has at least one of these is 0.90. The probability that he has both is

(a)0.25 (b)0.15 (c)0.255 (d)0.745

Solution :

P(B.com and CA)=p(A∩B)=p(A)+p(B)-p(AUB)=0.85+0.30-0.90=0.25

Answer : (a)

If a random variable takes values 0,1,2,3 with probabilities 0.4,0.3,0.2 and 0.1 then the expected value of x is

(a)0 (b)0.4 (c)1 ( d) (0.6)

Solution :

• Since ∑p(x)=1 , we have 0.3x1+0.2x2+0.1x3=1 • Answer : (c)

The probabilities that there is atleast one error in an account statement prepared by three persons A,B,C are 0.2,0.1,and 0.05. If A,B,C prepares 200,500,1000 such statements the expected number of total correct statements is (a)1460 (b)1470 (c)1590 (d)1560

Solution :

Ans : (d)

P(atleast one error)

P(No. of error)

No.of statements

xp

A 0.2 0.8 200 160 B 0.1 0.9 500 450 C 0.05 0.95 1000 950

E(x)=1560

A random variable X has the probability mass function X : 1 2 3 4 5 P(x) : 2k 3k 4k 3k^2 7k^2 Find p(x<2) (a)1/10 (b)1/3 (c)1/2 (d)3/10 Solution : Since ∑p(x)=1 we have 2k+3k+4k+3k^2+7k^2=1 10k^2+9k-1=0 K=1/10 , -1 Since k≠-1 we have k=1/10 P(x<2)=5k=5/10 = ½ Ans : (c)

For the probability distribution f(x)=1, 0<x<1

0 otherwise

The expected value of x is (a)∞ (b)-0.5 (c)1 (d)0.5

Solution :

E(x)=∫xf(x)dx=1

Ans : (c)

The random variable X has the following distribution X : -1 2 3 P(x) : 1/6 1/2 1/3 The value of E(3x^2-2x) is (a)28/6 (b)65/3 (c)56/3 (d)32/3 Solution : 3x^2-2x : 1 16 33 P(x) : 12/80 ½ 1/3 E(3x^2-2x)=56/3 Answer : (c)

A random variable x takes the values 1,1/2,1/4 with respective probabilities ¼,1/2,1/4 Then the value of E(1/x^2) is (a)25/4 (b)17/4 (c)9/4 (d)11/4

Solution :

X : 1 ½ ¼

X^2 : 1 ¼ 1/16

1/x^2 : 1 4 16

p(x) : ¼ ½ ¼

E(1/x2)=1/4 +4/2+4 = 25/4 Ans: (a)

The random variable X has the following distribution X : -2 3 1 P(x) : 1/3 1/2 1/6 The value of E(2x+5) is (a)6 (b)1 (c)7 (d)5 Solution : E(x)=-2/3+3/2+1/6 = 1 E(2x+5)=7 Ans: (c )

The random variable X has the following probability distribution

X : -2 3 1

P(x) : 1/3 1/2 1/6

The value of E(x^2) is

(a)4 (b)6 (c)-4 (d)-6

Solution :

X^2 : 4 9 1

P(x) : 1/3 ½ 1/6

E(x^2)=4/3+9/2+1/6=6

Ans : (b)

A bag contains 10 balls of which 3 are red. Five balls

are drawn from the bag . The probability that 3 of which are red is (a)1/6 (b)3/8 (c)1/12 (d)5/12

Solution : p(3 are red)= (3C3x7C2)/10C5 =1/12

Ans : (c)

A and B are two independent events with p(A)=1/3 , p(AuB)=4/5 Then p(B) is

(a)7/8 (b)9/10 (c)11/12 (d)5/6

Solution :

P(AuB)=p(A)+p(B)-p(A)p(B)

4/5 = 1/3+p(B)-1/3 p(B)

P(B)=7/8

Ans : (a)

The random variable X has the following probability distribution X : 0 1 2 3 P(x) : 0 k 2k 3k The n p(x<3) is (a)1/6 (b)1/3 (c)1/4 (d)1/2 Solution : Since ∑p(x)=1 , 6k=1 K=1/6 P(x<3)=3k=3/6=1/2 Ans : (d)

If the expected values of two random variables X and Y are 8 and 5 respectively the expected value of 2x-3y is

(a)0 (b)31 (c)1 (d)-10

Solution : E(2x-3y)=2E(x)-3E(y)=16-15=1

Ans : (c)

If the expected values of two independent random variables X and Y are 2 and 5 respectively the expected value of xy is

(a)15 (b)10 (c)7 (d)None of these

Solution : E(xy)=E(x)E(y)=2x5=10

Ans : (b)

The probabilities of A and B winning in a race are 1/6 and 1/3 respectively . The probability of A or B winning is (a)2/3 (b)4/9 (c)1/2 (d)5/6

Solution :

P(AorB)=p(A)+p(B)=1/6 +1/3=1/2

Ans : (c)

The probabilities of A,B,C winning a race are ½,1/4, 1/6

respectively. The probability one or others of A,B,C winning the race is (a)5/6 (b)2/3 (c)11/12 (d)3/4

Solution :

P(A)=1/2 p(B)=1/4 p(c)=1/6

P(Aor B or C )=1/2 + ¼ +1/6 =11/12

Ans : (c)

If A and B toss a die, A gets a prize of Rs 600 , if both the dice show the same number , otherwise B wins. The expected amount of B winning is

(a)400 (b)500 (c)410 (d)550

Solution :

Let X denote the amount of B winning .

E(x)=600x5/6+0x1/6=500

Ans : (b)

A business man ventures on a project in which he can make a profit of Rs3,00,000 or a loss Rs1,00,000.The probability of making a profit of Rs3,00,000 is 0.6.Then the expected gain or loss is

(a)Rs180,000 (b)Rs1,40,000 (c)2,00,000 (d)Rs60,000

Solution :

The expected gain = 300000x0.6-100000x0.4 = Rs1,40,000

Ans : (b)

The wages of eight workers in rupees are 62,50,70,40,56,45,45,32.If one of the worker is chosen at random the probability that his wage is above the average is (a)2/5 (b)3/8 (c)1/2 (d)5/8

Solution : Mean = (62+50+70+40+56+45+45+32)/8 =50

P(Mean>50)=3/8

Ans : (b)

If X and Y are independent random variables with standard deviation 3 and 4 then the standard deviation of (X+Y) is (a)7 (b)1 (c)25 (d)5

Solution : If X and Y are independent random variables the v(X+Y)=v(X)+v(Y)=9+16=25

Standard deviation of (X+Y) = √25=5

Ans : (d)