1. A college has 10 basketball players. A 5-member team and a
captain will be selected out of these 10 players. How many different selections
can be made?

a) 1260

b) 210

c) 210 × 6!

d) 1512

e) None of these

Answer: A)

A team of 6 members has to be
selected from the 10 players. This can be done in 10C6 or 210 ways.

Now, the captain can be selected from
these 6 players in 6 ways.

Therefore, total ways the selection
can be made is 210×6= 1260

2. From a group of 7 men and 6 women, five persons are to be
selected to form a committee so that at least 3 men are there on the committee.
In how many ways can it be done?

a) 564

b) 735

c) 756

d) 657

e) 854

Answer: C)

We may have (3 men and 2 women) or (4
men and 1 woman) or (5 men only).

Required number of ways= (^{7}C_{3}
× ^{6}C_{2}) + (^{7}C_{4} × ^{6}C_{1})
+ ^{7}C_{5} = 756.

3. Out of 7 consonants and 4 vowels, how many words of 3
consonants and 2 vowels can be formed?

a) 25200

b) 52000

c) 120

d) 24400

e) 12000

Answer: A)

Number of ways of selecting (3
consonants out of 7) and (2 vowels out of 4) = (^{7}C_{3}
× ^{4}C_{2}) = 210.

Number of groups, each having 3
consonants and 2 vowels = 210.

Each group contains 5 letters.

Number of ways of arranging 5
letters among themselves = 5! = 120

Required number of ways = (210 x 120)
= 25200.

4. In a group of 6 boys and 4 girls, four children are to be
selected. In how many different ways can they be selected such that at least
one boy should be there?

a) 209

b) 290

c) 200

d) 208

e) 280

Answer: A)

We may have (1 boy and 3 girls) or (2
boys and 2 girls) or (3 boys and 1 girl) or (4 boys).

Required number of ways = (^{6}C_{1}
× ^{4}C_{3}) + (^{6}C_{2} × ^{4}C_{2})
+ (^{6}C_{3} × ^{4}C_{1}) + ^{6}C_{4}
= 209.

5. A box contains 2 white balls, 3 black balls and 4 red
balls. In how many ways can 3 balls be drawn from the box, if at least one
black ball is to be included in the draw?

a) 48

b) 64

c) 63

d) 45

e) 52

Answer: B)

We may have (1 black and 2 non-black)
or (2 black and 1 non-black) or (3 black).

6. Tickets numbered 1 to 20 are mixed up and then a ticket is
drawn at random. What is the probability that the ticket drawn has a number
which is a multiple of 3 or 5?

a) ½

b) 3/5

c) 9/20

d) 8/15

e) 7/18

Answer: C)

Here, S = {1, 2, 3, 4, ...., 19, 20}.

Let E = event of getting a multiple
of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.

P(E) = n(E)/n(S) = 9/20.

7. In a box, there are 8 red, 7 blue and 6 green balls. One
ball is picked up randomly. What is the probability that it is neither red nor
green?

a) 1/3

b) 3/5

c) 8/21

d) 7/21

e) 9/21

Answer: A)

Total number of balls = (8 + 7 + 6) =
21.

Let E = event that the ball drawn is
neither red nor green = event that the ball drawn is blue.

n(E) = 7

P(E) = n(E)/n(S) = 7/21 = 1/3

8. A bag contains 2 red, 3 green and 2 blue balls. Two balls
are drawn at random. What is the probability that none of the balls drawn is
blue?

a) 10/21

b) 11/21

c) ½

d) 2/7

e) 3/7

Answer: A)

Total number of balls = (2 + 3 + 2) =
7

Let S be the sample space.

Then, n(S) = Number of ways of
drawing 2 balls out of 7 =7C2 = 21

Let E = Event of drawing 2 balls,
none of which is blue.

n(E) = Number of ways of drawing 2
balls out of (2 + 3) balls = 5C2 = 10

Therefore, P(E) = n(E)/n(S) = 10/ 21

9. In a class, there are 15 boys and 10 girls. Three students
are selected at random. The probability that 1 girl and 2 boys are selected,
is:

a) 21/46

b) 1/5

c) 3/25

d) 1/50

e) 2/25

Answer: A)

Let S be the sample space and E be
the event of selecting 1 girl and 2 boys.

Then, n(S) = Number ways of selecting
3 students out of 25 = 25C3 = 2300.

n(E)= 10C1 × 15C2 =
1050.

P(E) = n(E)/n(s) = 1050/2300 = 21/46

10. A problem is given to three students whose chances of
solving it are 1/2, 1/3 and 1/4 respectively. What is the probability that the
problem will be solved?