**1. In a simultaneous throw of two dice, find the probability of getting a total of 7.**

a) 2/3

b) 1/6

c) 2/7

d) 3/7

e) 4/7

**2. A coin is tossed successively three times. Find the probability of getting exactly one head or two heads.**

a) ¼

b) ½

c) ¾

d) 1

e) None of these

**3. Three unbiased coins are tossed. What is the probability of getting**

**(i) all heads (ii) two heads (iii) one head (iv) at least one head (v) at least two heads?**

a) 1/8, 3/8, 3/8, 7/8, ½

b) 3/8, 1/8, 3/8, 7/8, ½

c) 1/8, 7/8, 3/8, 3/8, ½

d) 1/8, 3/8, 3/8, 7/8, ¼

e) 1/8, 3/8, 3/8, ¼, ½

**4. What is the probability that a number selected from the numbers 1, 2, 3, ..., 24, 25, is a prime number, when each of the given numbers is equally likely to be selected?**

a) 7/25

b) 8/25

c) 9/25

d) 10/25

e) 11/25

**5. Tickets numbered from 1 to 20 are mixed up together and then a ticket is drawn at random. What is the probability that the ticket has a number which is a multiple of 3 or 7?**

a) 2/3

b) 2/7

c) 2/9

d) 2/5

e) 2/11

**6. One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that the card drawn is red?**

a) 1

b) ½

c) 1/3

d) ¼

e) None of these

**7. One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that the card drawn is king?**

a) ½

b) 1/13

c) 1/26

d) 7/13

e) 2/13

**8. One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that the card drawn is red and a king.**

a) ½

b) 1/13

c) 1/26

d) 7/13

e) 2/13

**9. One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that the card drawn is either red or a king.**

a) ½

b) 1/13

c) 1/26

d) 7/13

e) 2/13

**10. Two cards are drawn at random from a pack of 52 cards. What is the probability that the drawn cards are both aces?**

a) 2/221

b) 5/221

c) 1/221

d) 3/221

e) 7/221

**11. What is the probability that a leap year selected at random will contain 53 Sundays?**

a) 1/7

b) 2/7

c) 3/7

d) 4/7

e) 5/7

**12. A bag contains 9 black and 12 white balls. One ball is drawn at random. What is the probability that the ball drawn is black?**

a) 1/7

b) 2/7

c) 3/7

d) 4/7

e) 5/7

**13. A bag contains 9 black and 12 white balls. One ball is drawn at random. What is the probability that all the three balls are white?**

a) 5/143

b) 28/143

c) 40/143

d) Can’t be determined

e) None of these

**14. A bag contains 9 black and 12 white balls. One ball is drawn at random. What is the probability that all the three balls are red?**

a) 5/143

b) 28/143

c) 40/143

d) Can’t be determined

e) None of these

**15. A bag contains 9 black and 12 white balls. One ball is drawn at random. What is the probability that one ball is red and two balls are white?**

a) 5/143

b) 28/143

c) 40/143

d) Can’t be determined

e) None of these

**Solutions:**

**1. B)**We know that in a throw of two dice the total number of possible outcomes is (6 × 6 =) 36.

Thus, if S is the sample space, then
n(s) = 36

Let E be the event of getting a total
of 7.

Then, E = {(1, 6), (2, 5), (3, 4),
(4, 3), (5, 2), (6, 1)}

Thus, n(E) = 6

Therefore P(E) = n(E)/n(S) = 6/36 =
1/6

**2. C)**Let S be the sample space. Then,

S = {HHH,
HHT, HTH, THH, TTH, THT, HTT, TTT}

Let E be
the event of getting exactly one head or exactly two heads.

Then, E =
{HHT, HTH, THH, HTT, THT, T TH}

Clearly,
n(E) = 6 and n(S) = 8

Therefore, P(E) = n(E)/n(S) = 6/8 = ¾

**3. A)**Clearly, the sample space is S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}

Let E1, E2, E3, E4 and E5 be the
events of getting all heads; two heads; one head; at least one head and at
least two heads respectively.

Then, E1 = {HHH}; E2 = {HHT, HTH,
THH}; E3 = {HTT, THT, TTH}; E4 = {HTT, THT, TTH, HHT, HTH, THH, HHH} and E5 =
{HHT, HTH, THH, HHH}

By the formula P(E) = n(E)/n(S)

P(E1) = 1/8; P(E2) = 3/8; P(E3) = 3/8;
P(E4) = 7/8; P(E5) = 4/8 = ½

**4. C)**Sample space, S = {1, 2, 3, 4, ... 24, 25}

Let E be the event of selecting a
prime number then, E = {2, 3, 5, 7, 11, 13, 17, 19, 23}

Therefore P(E) = 7/25

**5. D)**Clearly, the sample space S has 20 points.

Let E be the event of getting a
multiple of 3 or 7.

Then E = {3, 6, 9, 12, 15, 18, 7, 14}

Therefore, P (a multiple of 3 or 7) =
8/20 = 2/5

**6. B)**Let S denote the sample space. Then n(S) = 52

Let E be the event of drawing a red
card.

Since the number of red cards is 26,
we have n(E) = 26

Therefore, P(a red card) = P(E) = 26/52
= ½

**7. B)**Let S denote the sample space. Then n(S) = 52

Let E be the event of drawing a king.

Since the number of kings is 4, we
have n(E) = 4

Therefore, P(E) = 4/52 = 1/13

**8. C)**Let S denote the sample space. Then n(S) = 52

Let E be the event of drawing a red
card which is a king. Since the number of red kings is 2, we have n(E) = 2

Therefore, P(E) = 2/52 = 1/26

**9. D)**Let S denote the sample space. Then n(S) = 52

Let E be the event of drawing a red
card or a king.

Clearly, there are 26 red cards
(including 2 red kings) and there are 2 more kings.

Thus, n(E4) = (26 + 2 =) 28

Therefore, P(E) = 28/52 = 7/13

**10. C)**Let S be the sample space. Then, n(S) = number of ways of selecting 2 cards out of 52 =

^{52}C

_{2}= (52 × 51)/2 = 1326

Let E be the event of getting both
the aces. Then, n(E) = number of ways of selecting aces out of 4 =

^{4}C_{2}= (4 × 3)/2 = 6
Therefore P(both aces) = 6/1326 =
1/221

**11. B)**A leap year contains 366 days and therefore 52 weeks and 2 days. Clearly, there are 52 Sundays in 52 weeks.

For the remaining 2 days, they may
be:

(i) Sunday and Monday

(ii) Monday and Tuesday

(iii) Tuesday and Wednesday

(iv) Wednesday and Thursday

(v) Thursday and Friday

(vi) Friday and Saturday

(vii) Saturday and Sunday

Now, for having 53 Sundays in the
year, one of the above 2 days must be Sunday. Thus, out of the above 7
possibilities, 2 favour the event that one of the two days is a Sunday.

Therefore required probability = 2/7

**12. C)**Total number of balls = (9 + 12 =) 21

Thus, if S is the sample space, then
n(S) =

^{21}C_{1}= 21
And, if E is the event of getting a
black ball, then n(E) =

^{9}C_{1}= 9
Therefore required probability
P(getting a black ball) = 9/21 = 3/7

**(13 – 15):**

Total number of balls = (8 + 5) = 13

Let S be the sample space. Then n(S)
= number of ways of selecting 3 balls out of 13 =

^{13}C_{3}= (13 × 12 × 11)/(3 × 2 × 1) = 286**13. A)**Let E be the event of getting 3 white balls. Then, n(E) = number of ways of selecting 3 balls out of 5 =

^{5}C

_{3}= 10

Therefore required probability =
10/286 = 5/143

**14. B)**Let E be the event of getting 3 red balls. Then, n(E) = number of ways of selecting 3 balls out of 8 =

^{8}C

_{3}= 56

Therefore required probability = 56/286
= 28/143

**15. C)**Let E be the event of getting 1 red and 2 white balls.

Then, n(E) = (number of ways of
selecting 1 ball out of 8) × (number of ways of selecting 2 balls out of 5)

= 8C1 × 5C2 = 8 × 10 = 80

Therefore, P(getting 1 red and 2
white balls) = 80/256 = 40/143