**1. ** (i) 1 (ii) 0, impossible event (iii) 1, sure or certain event

(iv) 1 (v) 0, 1

**2.** (i) In the experiment, “A driver attempts to start a car. The car starts or does not start”, we are not justified to assume that each outcome is as likely to occur as the other. Thus, the experiment has no equally likely outcomes.

(ii) In the experiment, “A player attempts to shoot a basket ball. She/he shoots or misses the shot”, we are not justified to assume that each outcome is as likely to occur as the other. Thus, the experiment has no equally likely outcomes.

(iii) In the experiment “A trial is made to answer a true-false question. The answer is right or ” We know, in advance, that the result can lead in one of the two possible ways -either right or wrong. We can reasonably assume that each outcome, right or wrong, is likely to occur as the other.

Thus, the outcomes right or wrong are equally likely.

(iv) In the experiment, “A baby is born, It is a boy or a girl”. We know, in advance that the outcome can lead in one of the two possible outcomes – either a boy or a girl. We are justified to assume that each outcome, boy or girl, is likely to occur as the other. Thus, the outcomes boy or girl are equally likely.

**3. ** The tossing of a coin is considered to be a fair way of deciding which team should get the ball at the beginning of a football game as we know that the tossing of the coin only land in one of two possible ways – either head up or tail up. It can reasonably be assumed that each outcome, head or tail, is as likely to occur as the other, i.e., the outcomes head and tail are equally likely. So the result of the tossing of a coin is completely unpredictable.

**4. (B) **Since the probability of an event E is a number P(E) such that

0 ≤ P(E) ≤ 1

∴ -1.5 cannot be the probability of an event.

**5.** Since P(E) + P (not E) = 1

∴ P (not E) = 1 – P(E) = 1 – 0.05 = 0.95

**6. ** (i) Consider the event related to the experiment of taking out of an orange flavoured candy from a bag containing only lemon flavoured candies.

Since no outcome gives an orange flavoured candy, therefore, it is an impossible event so its probability is 0.

(ii) Consider the event of taking a lemon flavoured candy out of a bag containing only lemon flavoured candies. This event is a certain event so its probability is 1.

**7.** Let E be the event of having the same birthday

⇒ P(E) = 0.992

**8. ** There are 3 + 5 = 8 balls in a bag. Out of these 8 balls, one can be chosen in 8 ways.

∴ Total number of elementary events = 8

(i) Since the bag contains 3 red balls, therefore, one red ball can be drawn in 3 ways.

∴ Favourable number of elementary events = 3

Hence P (getting a red ball) =

(ii) Since the bag contains 5 black balls along with 3 red balls, therefore one black (not red) ball can be drawn in 5 ways.

∴ Favourable number of elementary events = 5

Hence P (getting a black ball) =

**9. ** Total number of marbles in the box = 5 + 8 + 4 = 17 Total number of elementary events = 17

(i) There are 5 red marbles in the box.

∴ Favourable number of elementary events = 5

(ii) There are 8 white marbles in the box.

∴ Favourable number of elementary events = 8

∴ P (getting a white marble) =

There are 5 + 8 = 13 marbles in the box, which are not green.

∴Favourable number of elementary events = 13

∴ P (not getting a green marble) =

**10.** Total number of coins in a piggy bank = 100 + 50 + 20 + 10 = 180

∴ Total number of elementary events = 180

(i) There are one hundred 50 coins in the piggy bank.

∴ Favourable number of elementary events = 100

∴ P (falling out of a 50 p coin) =

(ii) There are 100 + 50 + 20 = 170 coins other than Rs. 5 coin.

∴ Favourable number of elementary events = 170

∴ P (falling out of a coin other than Rs. 5 coin) =

**11.** Total number of fish in the tank = 5 + 8 = 13

∴ Total number of elementary events = 13

There are 5 male fishes in the tank.

∴ Favourable number of elementary events = 5

Hence, P (taking out a male fish) =

**12.** Out of 8 numbers, an arrow can point any of the numbers in 8 ways.

∴ Total number of favourable outcomes = 8

(i) Favourable number of outcomes = 1

Hence, P (arrow points at 8) =

(ii) Favourable number of outcomes = 4

Hence, P (arrow points at an odd number) =

(iii) Favourable number of outcomes = 6

Hence, P (arrow points at a number > 2) =

(iv) Favourable number of outcomes = 8

Hence, P (arrow points at a number < 9) =

**13.** Total number of favourable outcomes of throwing a dice = 6

(i) On a dice, the prime numbers are 2, 3 and 5. Therefore favourable outcomes = 3

Hence P (getting a prime number) =

(ii) On a dice, the number lying between 2 and 6 are 3, 4, 5. Therefore favourable outcomes = 3

Hence P (getting a number lying between 2 and 6) =

(iii) On a dice, the odd numbers are 1, 3 and 5. Therefore favourable outcomes = 3

Hence P (getting an odd number) =

**14.** Total number of favourable outcomes = 52

(i) There are two suits of red cards, i.e., diamond and heart. Each suit contains one king.

∴ Favourable outcomes = 1

Hence, P (a king of red colour) =

(ii) There are 12 face cards in a pack.

∴ Favourable outcomes = 12

(iii) There are two suits of red cards, i.e., diamond and heart. Each suit contains 3 face cards.

∴ Favourable outcomes = 2 x 3 = 6

(iv) There are only one jack of heart.

∴ Favourable outcome = 1

Hence, P (the jack of hearts) =

(v) There are 13 cards of spade.

∴ Favourable outcomes = 13

(vi) There is only one queen of diamonds.

∴ Favourable outcome = 1

Hence, P (the queen of diamonds) =

**15. ** Total number of favourable outcomes = 5

(i) There is only one queen.

∴ Favourable outcome = 1

(ii) In this situation, total number of favourable outcomes = 4

(a) Favourable outcome = 1

(b) There is no card as queen.

∴ Favourable outcome = 0

**16.** Total number of favourable outcomes = 132 + 12 = 144

Number of favourable outcomes = 132

Hence, P (getting a good pen) =

**17.** (i) Total number of favourable outcomes = 20 Number of favourable outcomes = 4

Hence P (getting a defective bulb) =

(ii) Now total number of favourable outcomes = 20 – 1 = 19

Number of favourable outcomes = 19 – 4 = 15

Hence P (getting a non-defective bulb) =

**18. ** Total number of favourable outcomes = 90

(i) Number of two-digit numbers from 1 to 90 are 90 – 9 = 81

∴ Favourable outcomes = 81

Hence, P (getting a disc bearing a two-digit number) =

(ii) From 1 to 90, the perfect squares are 1, 4, 9, 16, 25, 36, 49, 64 and 81.

∴ Favourable outcomes = 9

Hence P (getting a perfect square) =

(iii) The numbers divisible by 5 from 1 to 90 are 18.

∴ Favourable outcomes = 18

Hence P (getting a number divisible by 5) =

**19.** Total number of favourable outcomes = 6

(i) Number of favourable outcomes = 2

Hence, P (getting a letter A) =

(ii) Number of favourable outcomes = 1

Hence P (getting a letter D) =

**20. **Total area of the given figure (rectangle) = 3 x 2 = 6 m^{2}

Hence, P (die to land inside the circle) =

**21. ** Total number of favourable outcomes = 144

Number of non-defective pens = 144 – 20 = 124

∴ Number of favourable outcomes = 124

Hence P (she will buy) = P (a non-defective pen) =

Number of favourable outcomes = 20

Hence P (she will not buy) = P (a defective pen) =

**22.** Total favourable outcomes of throwing two dice are:

(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

∴ Total number of favourable outcomes = 36

(i) Favourable outcomes of getting the sum as 3 = 2

Hence P (getting the sum as 3) =

Favourable outcomes of getting the sum as 4 = 3

Hence P (getting the sum as 4) =

Favourable outcomes of getting the sum as 5 = 4

Hence P (getting the sum as 5) =

Favourable outcomes of getting the sum as 6 = 5

Hence P (getting the sum as 6) =

Favourable outcomes of getting the sum as 7 = 6

Hence P (getting the sum as 7) =

Favourable outcomes of getting the sum as 9 = 4

Hence P (getting the sum as 9) =

Favourable outcomes of getting the sum as 10 = 3

Hence P (getting the sum as 10) =

Favourable outcomes of getting the sum as 11 = 2

Hence P (getting the sum as 11) =

(ii) I do not agree with the argument given here. Justification has already been given in part (i).

**23.** The outcomes associated with the experiment in which a coin is tossed thrice:

HHH, HHT, HTH, THH, TTH, HTT, THT, TTT

Therefore, Total number of favourable outcomes = 8

Number of favourable outcomes = 6

**24. **(i) The outcomes associated with the experiment in which a dice is thrown is twice:

(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

Therefore, Total number of favourable outcomes = 36 Now consider the following events:

A = first throe shows 5 and B = second throw shows 5

Therefore, the number of favourable outcomes = 6 in each case

(ii) Let S be the sample space associated with the random experiment of throwing a die twice. Then, n(S) = 36

∴ A ∩ B = first and second throw shoe 5, i.e. getting 5 in each throw.

We have, A = (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

And B = (1, 5) (2, 5) (3, 5) (4, 5) (5, 5) (6, 5)

∴ Required probability = Probability that at least one of the two throws shows 5

**25. ** (i) **Incorrect: **We can classify the outcomes like this but they are not then, ‘equally likely’. Reason is that ‘one of each’ can result in two ways – from a head on first coin and tail on the second coin or from a tail on the first coin and head on the second coin. This makes it twice as likely as two heads (or two tails).

(ii) **Correct: **The two outcomes considered in the question are equally likely.

**26.** Total favourable outcomes associated to the random experiment of visiting a particular shop in the same week (Tuesday to Saturday) by two customers Shyam and Exta are:

(T, T) (T, W) (T, TH) (T, F) (T, S)

(W, T) (W, W) (W, TH) (W, F) (W, S)

(TH, T) (TH, W) (TH, TH) (TH, F) (TH, S)

(F, T) (F, W) (F, TH) (F, F) (F, S)

(S, T) (S, W) (S, TH) (S, F) (S< S)

∴ Total number of favourable outcomes = 25

(i) The favourable outcomes of visiting on the same day are (T, T), (W, W), (TH, TH), (F, F) and (S, S).

∴ Number of favourable outcomes = 5

(ii) The favourable outcomes of visiting on consecutive days are (T, W), (W, T), (W, TH), (TH, W), (TH, F), (F, TH), (S, F) and (F, S).

∴ Number of favourable outcomes = 8

(iii) Number of favourable outcomes of visiting on different days are 25 – 5 = 20

∴ Number of favourable outcomes = 20

**27. ** Complete table is as under:

It is clear that total number of favourable outcomes = 6 x 6 = 36

(i) Number of favourable outcomes of getting total score even are 18

Hence P (getting total score even) =

(ii) Number of favourable outcomes of getting total score 6 are 4

Hence P (getting total score 6) =

(iii) Number of favourable outcomes of getting total score at least 6 are 15

Hence P (getting total score at least 6) =

**28.** Let there be x blue balls in the bag.

Total number of balls in the bag = 5+x

Now, P_{1} = Probability of drawing a blue ball =

And P_{2} = Probability of drawing a blue ball =

But according to question, P_{1} = 2P_{2}

Hence, there are 10 blue balls in the bag.

**29. ** There are 12 balls in the box.

Therefore, total number of favourbale outcomes = 12

The number of favourable outcomes = x

Therefore P_{1} = P (getting a black ball) =

If 6 more balls put in the box, then

Total number of favourable outcomes = 12 + 6 = 18

And Number of favourable outcomes = x + 6

P_{2} = P (getting a black ball) =

According to question, P_{2} = 2P_{1}

**30.** Here, Total number of favourable outcomes = 24

Let there be x green marbles.

Therefore Favourable number of outcomes = x

Therefore number of green marbles are 16

And number of blue marbles = 24 – 16 = 8.