1.A bag contains 1100 tickets numbered 1, 2, 3, … 1100. If a ticket is drawn out of it at random, what is the probability that the ticket drawn has the digit 2 appearing on it?

a. 291/1100

b. 292/1100

c. 290/1100

d. 301/1100

Answer: c

Explanation:

Numbers which dont have 2 from 1 to 9 = 8

Numbers which don’t have 2 from 10 to 99:

Let us take two places _ _. Now left most place is fixed in 8 ways. Units place is filled with 9 ways. Total 72 numbers.

Numbers which don’t have 2 from 100 to 999 =_ _ _ = 8 × 9 × 9 = 648

Numbers which don’t have 2 from 1000 to 1099 =10_ _ = 9 × 9 = 81

Finally 1100 does not have 2. So 1.

Total number with no 2 in them = 8 + 72 + 648 + 81 + 1= 810

Tickets with 2 in them = 1100 – 810 = 290

Required probability = 290 / 1100

2. In how many ways a team of 11 must be selected a team 5 men and 11 women such that the team must comprise of not more than 3 men.

a. 1565

b. 2256

c. 2456

d. 1243

Answer: b

Explanation:

Maximum 3 men can be played which means there can be 0, 1, 2, 3 men in the team.

(5C0×11C11)+(5C1×11C10)(5C0×11C11)+(5C1×11C10) + (5C2×11C9)+(5C3×11C8)(5C2×11C9)+(5C3×11C8) = 2256

3.Four people each roll a four die once. Find the probability that at least two people will roll the same number ?

a. 5/18

b. 13/18

c. None of the given choices

d. 1295/1296

Answer: b

Explanation:

The number of ways of rolling a dice where no two numbers probability that no one rolls the same number = 6 x 5 x 4 x3

Now total possibilities of rolling a dice = 6^4

The probability that a no one gets the same number =

(6*5*4*3)/6^4 = 5/18

So the probability that at least two people gets same number =1-5/18 = 13/18

4.In a single throw with two dice, find the probability that their sum is a multiple either of 3 or 4.

a. 1/3

b. 1/2

c. 5/9

d. 17/36

Answer: Their sum can be 3,4,6,8,9,12

Explanation: For two dice, any number from 2 to 7 can be get in (n-1) ways and any number from 8 to 12 can be get in (13 – n) ways.

Then possible ways are 2 + 3 + 5 + 5 + 4 + 1 = 20 possible cases.

So probability is (20/36)=(5/9)

5.A man who goes to work long before sunrise every morning gets dressed in the dark. In his sock drawer he has 6 black and 8 blue socks. What is the probability that his first pick was a black sock, but his second pick was a blue sock?

a.24/71

b.22/91

c.24/91

d.22/71

Answer:c

Explaination:This is a case of without replacement. We have to multiply two probabilities. 1. Probability of picking up a black sock, and probability of picking a blue sock, given that first sock is black.

6C1/14C1 × 8C1/1C1 =24/91.

6.There are 6 red balls,8 blue balls and 7 green balls in a bag. If 5 are drawn with replacement, what is the probability at least three are red?

a.312/16807

b.314/16907

c.322/13904

d.331/19807

Answer:

Explanation:At least 3 reds means we get either : 3 red or 4 red or 5 red. And this is a case of replacement.

case 1 : 3 red balls : 6/21 x 6/21 x 6/21 x 15/21 x 15/21

case 2 : 4 red balls : 6/21 x 6/21 x 6/21 x 6/21 x 15/21

case 3 : 5 red balls : 6/21 x 6/21 x 6/21 x 6/21 x 6/21

Total probability = = (6/21 x 6/21 x 6/21 x 15/21 x 15/21)+(6/21 x 6/21 x 6/21 x 6/21 x (15 )/21)+ (6/21 x 6/21 x 6/21 x 6/21 x 6/21)

= 312/16807

7.2/3rd of the balls in a bag are blue, the rest are pink. if 5/9th of the blue balls and 7/8th of the pink balls are defective, find the total number of balls in the bag given that the number of non defective balls is 146.

a.431

b.24

c.221

d.23

Answer:c

Explanation: let the total balls be x 2/3x=blue & 1/3x=pink non defective balls=146 total no. of balls=defective+non defective hence defective=total-non defective(146) (2/3x)*5/9+(1/3x)*7/8=t-146 0.66203x=t-146 146=0.3303x x=432

8.From a bag containing 8 green and 5 red balls, three are drawn one after the other .the probability of all three balls beings green if the balls drawn are replaced before the next ball pick and the balls drawn are not replaced, are respectively?

a.512/2197, 336/2197

b.512/2197, 336/1716

c.336/2197, 512/2197

d.336/1716, 512/1716

Answer:a

Explanation:

The Probabilities of getting with replacement=8/13*8/13*8/13=512/2197 The Probabilities of getting without replacement =8/13*7/12*6/11=336/2197

9.There are two bags, one of which contains 5 red and 7 white balls and the other 3 red and 12 white balls. A ball is to be drawn from one or other of the two bags ; find the chances of drawing a red ball.

a.55/102

b.17/21

c.37/120

d.7/8

Answer:c

Explanation:

P(Getting a red ball from the first bag): (1/2)*(5/12) P(Getting a red ball from the 2nd bag): (1/2)*(3/15) P(Drawing a red ball): (1/2)*(5/12)+(1/2)*(3/15)= 37/120 (Ans).

10.3 dice are rolled. What is the probability that you will get the sum of the no’s as 10?

a.27/216

b.25/216

c.10/216

d.1/8

Answer: d

Explanation:

total events = 6*6*6=216 possible cases for sum equal to 10 are (1,3,6)-6 combinations (1,4,5)-6 combinations (2,3,5)-6 combinations (2,4,4)-3 combinations (3,3,4)-3 combinations (2,2,6)-3 combinations so total combinations are 27 so probability will be 27/216 = 1/8