NCERT Solutions for Exercise 15.1 Class 10 Maths Chapter 15 - Probability

Q1 (i) Complete the following statements: Probability of an event E + Probability of the event ‘not E’ = ________.

Answer:

Probability of an event E + Probability of the event ‘not E’ = 1. Because event E and event 'Not E' are both complementary.

Q1 (ii) The probability of an event that cannot happen is ______. Such an event is called ______.

Answer:

The probability of an event that cannot happen is 0. Such an event is called an impossible event.

Q1 (iii) The probability of an event that is certain to happen is_____ . Such an event is called _______.

Answer:

The probability of an event that is certain to happen is 1. Such an event is called a sure/certain event.

Q1 (iv) The sum of the probabilities of all the elementary events of an experiment is _______.

Answer:

The sum of the probabilities of all the elementary events of an experiment is 1.

Q1 (v) The probability of an event is greater than or equal to and less than or equal to ________ .

Answer:

The probability of an event is greater than or equal to 0 and less than or equal to 1.

Q2 (i) Which of the following experiments have equally likely outcomes? Explain. A driver attempts to start a car. The car starts or does not start.

Answer:

This statement does not have equally likely consequences, because the car may or may not start based on a variety of conditions, including fuel.

Q2 (ii) Which of the following experiments have equally likely outcomes? Explain: A player attempts to shoot a basketball. She/he shoots or misses the shot.

Answer:

It is not an equally likely event because it is determined by the player's talent and level of practice. If he is a professional player, he is more likely to make a good shot. Whereas an amateur player is more likely to miss the shot.

Q2 (iii) Which of the following experiments have equally likely outcomes? Explain: A trial is made to answer a true-false question. The answer is right or wrong.

Answer:

It is an equally likely event. The only options are true or false, and only one of them is correct.

Q2 (iv) Which of the following experiments have equally likely outcomes? Explain: A baby is born. It is a boy or a girl.

Answer:

It is an equally likely event. The only possibilities of gender are boy and girl. Hence, if not a boy, then a girl and vice versa.

Q3 Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?

Answer:

The answer becomes unbiased by using a coin toss due to its two possible results being heads or tails. The event that results in head or tail has equal odds since these two outcomes are equally likely outcomes, which makes tossing an unpredictable process deemed entirely unbiased.

Q4 Which of the following cannot be the probability of an event?

(A) 2/3

(B) –1.5

(C) 15%

(D) 0.7

Answer:

The probability of an event always lies between 0 and 1, and can be 0 or 1. Hence, the probability of an event can never be negative.

Therefore, (B) $-1.5$ cannot be the probability of an event. Also, Options (A), (C), and (D) all lie between 0 and 1; therefore, they can be the probability of an event.

Q5 If P(E) = 0.05, what is the probability of ‘not E’?

Answer:

We know,

$P(not\ E) + P(E)$ = 1

Given, $P(E)=0.05$

Therefore $P(notE)=1-0.05=0.95$

Thus, the probability of 'not E' is 0.95

Q6 (i) A bag contains lemon-flavored candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out- (i) an orange-flavored candy?

Answer:

According to the question, the bag solely contains lemon-flavoured candy. It does not include any orange-flavoured candy. As a result, every time, just lemon-flavoured candy will appear. Thus, $P(anorange-flavouredcandy)=0$.

Q6 (ii) A bag contains lemon-flavored candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out (ii) a lemon-flavored candy?

Answer:

As per the question, the bag contains only lemon-flavoured candies. So the event that is happening in this case is a sure event. Therefore, $P(alemonflavouredcandy)=1$

Q7 It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?

Answer:

Probability of two students not having the same birthday $P(\bar{E})=0.992$

Therefore, Probability of two students having the same birthday = $P(E)=1-P(\bar{E})$

$=1-0.992=0.008$

Thus, the probability that the 2 students have the same birthday is 0.008

Q8 (i) A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red?

Answer:

Given, the total number of balls in the bag = 8

Number of red balls = 3

Number of black balls = 5.

Let E be the event of getting a red ball

Therefore, Favourable outcomes = Number of red balls = 3

And, Total outcomes = Number of total balls = 8

Therefore,

Probability of the ball being drawn is red = $P=\frac{favourableoutcomes)}{Totaloutcomes}$ = $=\frac{3}{8}$

Q8 (ii) A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is not red?

Answer:

Total number of balls in the bag = 8

Number of red balls = 3

Number of black balls = 5

We know,

$P(notE)=P(\bar{E})=1-P(E)$

where $Eand\bar{E}$ are complementary events.

Therefore,

Probability of not getting the red ball $=1-Probabilityofgettingaredball$ = $$\begin{gathered} =1-\frac{3}{8} \\ =\frac{5}{8} \end{gathered}$$

Q9 (i) A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red?

Answer:

Total number of balls in the bag = 5 + 8 + 4 = 17

Let R be the event that the ball taken out is red

Thus, the total number of possible outcomes = 17

And, the total number of favourable outcomes = 5

Therefore $P(R)=\frac{favourableoutcomes}{totaloutcomes}=\frac{5}{17}$

Q9 (ii) A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (ii) white?

Answer:

Total number of balls in the bag = 5 + 8 + 4 = 17

Let W be the event that the ball taken out is white

Thus, the total number of possible outcomes = 17

And, the total number of favourable outcomes = 8

Therefore $P(W)=\frac{favourableoutcomes}{totaloutcomes}=\frac{8}{17}$

Q9 (iii) A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be not (iii) green?

Answer:

Total number of balls in the bag = 5 + 8 + 4 = 17

Let G be the event that the ball taken out is green

Thus, the total number of possible outcomes = 17

And, the total number of favourable outcomes = 4

Therefore $P(G)=\frac{favourableoutcomes}{totaloutcomes}=\frac{4}{17}$

$$\begin{gathered} \therefore P(notG)=P(\bar{G})=1-P(G) \\ =1-\frac{4}{17} \\ =\frac{13}{17} \end{gathered}$$

Thus, the required probability of not getting a green ball is $\frac{13}{17}$

Q10 (i) A piggy bank contains hundred 50p coins, fifty Rs 1 coins, twenty Rs 2 coins and ten Rs 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin will be a 50 p coin?

Answer:

Total number of coins in the piggy bank = 100 + 50 + 20 + 10 = 180

Let E be the event of getting a 50p coin.

Thus, the total number of possible outcomes = 180

And, the total number of favourable outcomes = 100

Therefore P(E) = \frac{favourable\ outcomes}{total\ outcomes} = \frac{100}{180}$ = $=\frac{5}{9}$

Thus, the probability of getting a 50p coin is $\frac{5}{9}$

Q10 (ii) A piggy bank contains hundred 50p coins, fifty Rs 1 coins, twenty Rs 2 coins and ten Rs 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin will not be a Rs 5 coin?

Answer:

Total number of coins in the piggy bank = 100 + 50 + 20 + 10 = 180

Let F be the event of getting an Rs. 5 coin.

Thus, the total number of possible outcomes = 180

And, the total number of favourable outcomes = 10

Therefore $P(F)=\frac{favourableoutcomes}{totaloutcomes}=\frac{10}{180}$

$=\frac{1}{18}$

Therefore $P(notgettingaRs.5coin)=P(\bar{F})$

$=1-P(F)=1-\frac{1}{18}=\frac{17}{18}$

Thus, the probability of not getting an Rs. 5 coin is $\frac{17}{18}$

Q11 Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish (see Fig. 15.4). What is the probability that the fish taken out is a male fish?

Answer:

Total number of fishes in the tank = 5 (male) + 8 (female) = 13

Let E be the event that the fish taken out is a male fish.

Thus, the total number of possible outcomes = 13

And, the total number of favourable outcomes = 5

Therefore $P(E)=\frac{favourableoutcomes}{totaloutcomes}=\frac{5}{13}$

Thus, the probability that the fish are taken out is a male fish is $\frac{5}{13}$

Q12 (i) A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see Fig. 15. 5 ), and these are equally likely outcomes. What is the probability that it will point at 8?

Answer:

Total possible outcomes = {1, 2, 3, 4, 5, 6, 7, 8}

Therefore, the number of possible outcomes = 8

Let E be the event of getting 8, and we can see the favourable outcome would be 1 (as 8 can appear only once)

Therefore $P(E)=\frac{favourableoutcomes}{totaloutcomes}=\frac{1}{8}$

Thus, the probability that it will point at 8 is $\frac{1}{8}$

Q12 (ii) A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see Fig. 15. 5 ), and these are equally likely outcomes. What is the probability that it will point at an odd number?

Answer:

Total posible outcomes = {1,2,3,4,5,6,7,8}

Therefore, the number of possible outcomes = 8

Let E be the event of pointing at an odd number.

And, the total number of odd numbers (favourable outcomes) = n({1,3,5,7}) = 4

Therefore $P(E)=\frac{favourableoutcomes}{totaloutcomes}=\frac{4}{8}$ or $=\frac{1}{2}$

Therefore, the probability of getting an odd number is $\frac{1}{2}$

Q12 (iii) A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see Fig. 15. 5 ), and these are equally likely outcomes. What is the probability that it will point at a number greater than 2?

Answer:

Total posible outcomes = {1,2,3,4,5,6,7,8}

Therefore, the number of possible outcomes = 8

Let E be the event of pointing at a number greater than 2

Thus, number of favouable outcomes= n({3,4,5,6,7}) = 5

Therefore $P(E)=\frac{favourableoutcomes}{totaloutcomes}=\frac{5}{8}$

Hence, the probability of pointing at a number greater than 2 is $\frac{5}{8}$

Q12 (iv) A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see Fig. 15. 5 ), and these are equally likely outcomes. What is the probability that it will point at a number less than 9?

Answer:

Total possible outcomes = {1,2,3,4,5,6,7,8}

Therefore, the number of possible outcomes = 8

Let E be the event of pointing at a number less than 9

Since all the numbers on the wheel are less than 9, this is a sure event.

Thus, the number of favourable outcomes = 8

Therefore $P(E)=\frac{favourableoutcomes}{totaloutcomes}=\frac{8}{8}=1$

Hence, the probability of pointing at a number less than 9 is $1$ .

Q13 (i) A die is thrown once. Find the probability of getting a prime number

Answer:

Possible outcomes when a die is thrown = {1,2,3,4,5,6}

Therefore, the number of possible outcomes = 6

Let E be the event of getting a prime number.

Prime numbers on the die are = {2,3,5}

Thus, the number of favourable outcomes = 3

Therefore $P(E)=\frac{favourableoutcomes}{totaloutcomes}=\frac{3}{6}$ or $=\frac{1}{2}$

Hence, the probability of getting a prime number is $\frac{1}{2}$

Q13 (ii) A die is thrown once. Find the probability of getting a number lying between 2 and 6

Answer:

Possible outcomes when a die is thrown once = {1,2,3,4,5,6}

Therefore, the number of possible outcomes = 6

Let F be the event of getting a number lying between 2 and 6

Numbers lying between 2 and 6 on the die are = {3,4,5}

Therefore, number of favourable outcomes = 3

Therefore $P(F)=\frac{favourableoutcomes}{totaloutcomes}=\frac{3}{6}$ or $=\frac{1}{2}$

Hence, the probability of getting a number lying between 2 and 6 is $\frac{1}{2}$

Q13 (iii) A die is thrown once. Find the probability of getting an odd number.

Answer:

Possible outcomes when a die is thrown = {1,2,3,4,5,6}

Therefore, the number of possible outcomes = 6

Let O be the event of getting an odd number.

Odd numbers on the die are = {1,3,5}

Thus, the number of favourable outcomes = 3

Therefore $P(O)=\frac{favourableoutcomes}{totaloutcomes}=\frac{3}{6}$ or $=\frac{1}{2}$

Hence, the probability of getting an odd number is $\frac{1}{2}$ .

Q14 (i) One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting a king of red colour.

Answer:

Total number of cards in a well-shuffled deck = 52

Hence, the total possible outcomes = 52

Let E be the event of getting a king of red colour.

There are only red colour kings: Hearts and diamonds

Hence, the number of favourable outcomes = 2

Therefore $P(E)=\frac{favourableoutcomes}{totaloutcomes}=\frac{2}{52}$ or $=\frac{1}{26}$

Thus, the probability of getting a king of red colour is $\frac{1}{26}$

Q14 (ii) One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting a face card

Answer:

Total number of cards in a well-shuffled deck = 52

Hence, the total possible outcomes = 52

Let E be the event of getting a face card.

Face cards: (J, Q, K) of each four suits.

Hence, the number of favourable outcomes = 3 $\times$ 4 = 12

Therefore $P(E)=\frac{favourableoutcomes}{totaloutcomes}=\frac{12}{52}$ or $=\frac{3}{13}$

Thus, the probability of getting a face card is $\frac{3}{13}$

Q14 (iii) One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting a red face card

Answer:

Total number of cards in a well-shuffled deck = 52

Hence, the total possible outcomes = 52

Let E be the event of getting a red face card.

Face cards: (J, Q, K) of hearts and diamonds

Hence, the number of favourable outcomes = 3 $\times$ 2 = 6

Therefore $P(E)=\frac{favourableoutcomes}{totaloutcomes}=\frac{6}{52}$ or $=\frac{3}{26}$

Therefore, the probability of getting a red face card is $\frac{3}{26}$

14 (iv) One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting the jack of hearts

Answer:

Total number of cards in a well-shuffled deck = 52

Hence, the total possible outcomes = 52

Let E be the event of getting the jack of hearts, and the number of times it is possible is 1.

Hence, the number of favourable outcomes = 1

Therefore $P(E)=\frac{favourableoutcomes}{totaloutcomes}=\frac{1}{52}$

Thus, the probability of getting the jack of hearts is $\frac{1}{52}$

14 (v) One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting a spade

Answer:

Total number of cards in a well-shuffled deck = 52

Hence, the total possible outcomes = 52

Let E be the event of getting a spade.

There are 13 cards in each suit. {2, 3, 4, 5, 6, 7, 8, 9,10, J, Q, K, A}

Hence, the number of favourable outcomes = 13

Therefore $P(E)=\frac{favourableoutcomes}{totaloutcomes}=\frac{13}{52}$ or $=\frac{1}{4}$

Therefore, the probability of getting a spade is $\frac{1}{4}$

14 (vi) One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting the queen of diamonds

Answer:

Total number of cards in a well-shuffled deck = 52

Hence, the total possible outcomes = 52

Let E be the event of getting the queen of diamonds

Hence, the number of favourable outcomes = 1

Therefore $P(E)=\frac{favourableoutcomes}{totaloutcomes}=\frac{1}{52}$

Therefore, the probability of getting the queen of diamonds is $\frac{1}{52}$

15 (i) Five cards—the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random. What is the probability that the card is the queen?

Answer:

Total number of cards = 5

Hence, the total possible outcomes = 5

There is only one queen.

Hence, favourable outcome = 1

Therefore $P(gettingaqueen)=\frac{1}{5}$

15 (ii) Five cards—the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random. If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace?

Answer:

When the queen is kept aside, there are only 4 cards left

Hence, the total possible outcomes = 4

There is only one ace.

Hence, favourable outcome = 1

Therefore $P(gettinganace)=\frac{1}{4}$

Thus, the probability of getting an ace is 0.25

15 (ii) Five cards—the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random. If the queen is drawn and put aside, what is the probability that the second card picked up is (b) a queen?

Answer:

When the queen is kept aside, there are only 4 cards left

Hence, the total possible outcomes = 4

Since there is no queen left.

Hence, favourable outcome = 0

Therefore $P(gettingaqueen)=\frac{0}{4}=0$

Thus, the probability of getting a queen is 0. Thus, it is an impossible event.

Q16 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen is taken out is a good one.

Answer:

Total number of pens = 132(good) + 12(defective)

Hence, the total possible outcomes = 144

The total number of good pens = number of favourable outcomes = 132

Therefore $P(gettingagoodpen)=\frac{favourableoutcome}{totaloutcome}=\frac{132}{144}=\frac{11}{12}$

17 (i) A lot of 20 bulbs contains 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?

Answer:

Total number of bulbs = 20

Hence, the total possible outcomes = 20

Number of defective bulbs = 4

Hence, the number of favourable outcomes = 4

Therefore $P(gettingadefectivebulb)=\frac{favourableoutcomes}{totaloutcomes}=\frac{4}{20}$ $=\frac{1}{5}$

Q17 (ii) Suppose the bulb is drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

Answer:

Total number of bulbs = 20

Hence, the total possible outcomes = 20

Number of defective bulbs = 4

Hence, the number of favourable outcomes = 4

Therefore $P(gettingadefectivebulb)=\frac{favourableoutcomes}{totaloutcomes}=\frac{4}{20}$ or $=\frac{1}{5}$

Thus, $P(gettinganondefectivebulb)=1-\frac{1}{5}=\frac{4}{5}$

Q18 (i) A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears a two-digit number

Answer:

Total number of discs = 90

Number of discs having a two-digit number between 1 and 90 (favourable outcomes) = 81

Therefore $P(gettingatwo-digitnumber)=\frac{favourableoutcomes}{totaloutcomes}=\frac{81}{90}$ or $=\frac{9}{10}$

Q18 (ii) A box contains 90 discs, which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears a perfect square number

Answer:

Total number of discs = 90

Perfect square numbers between 1 and 90 are {1, 4, 9, 16, 25, 36, 49, 64, 81}

Therefore, the total number of discs having perfect squares = 9.

Therefore $P(gettingaperfectsquare)=\frac{favourableoutcomes}{totaloutcomes}=\frac{9}{90}$ $=\frac{1}{10}$

Q18 (iii) A box contains 90 discs, which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears a number divisible by 5.

Answer:

Total number of discs = 90

Numbers between 1 and 90 that are divisible by 5 are {5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90}

Therefore, the total number of discs having numbers that are divisible by 5 = 18.

$\therefore P(gettinganumberdivisibleby5)=\frac{favourableoutcomes}{totaloutcomes}=\frac{18}{90}$

$=\frac{1}{5}$

Q19 (i) A child has a die whose six faces show the letters as given below:

The die is thrown once. What is the probability of getting (i) A?

Answer:

The six faces of the die contain: {A, B, C, D, E, A}

Thus, the total number of letters = 6

Since there are two A's,

Therefore, the number of favourable outcomes = 2

Therefore $P(gettingA)=\frac{favourableoutcomes}{totaloutcomes}=\frac{2}{6}$ $=\frac{1}{3}$

Hence, the probability of getting A is $\frac{1}{3}$

Q19 (ii) A child has a die whose six faces show the letters as given below:

The die is thrown once. What is the probability of getting (ii) D?

Answer:

The six faces of the die contain: {A, B, C, D, E, A}

Therefore, the total number of letters = 6

Since there is only one D,

Thus, the number of favourable outcomes = 1

Therefore $P(gettingA)=\frac{favourableoutcomes}{totaloutcomes}=\frac{1}{6}$

Hence, the probability of getting D is $\frac{1}{6}$

Q20 Suppose you drop a die at random on the rectangular region shown in Fig. 15.6. What is the probability that it will land inside the circle with a diameter of 1m?

Answer:

Here, the Total outcome in the area of the rectangle and the favourable outcome is the area of the circle.

Area of the rectangle = $l\times b=3\times 2=6m^2$

Area of the circle = $\pi r^2=\pi {\left(\frac{1}{2}\right)}^2=\frac{\pi }{4}{m}^2$

Therefore $P(diewilllandinsidethehole)=\frac{Areaofcircle}{Totalarea}$

$=\frac{\frac{\pi }{4}}{6}=\frac{\pi }{24}$

Q21 (i) A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that (i) She will buy it?

Answer:

Total number of pens = 144

Total number of defective pens = 20

Therefore, the number of good pens = 144 - 20 = 124

She will buy it if the pen is good.

Therefore, the probability that she buys = the probability that the pen is good =

$P(gettingagoodpen)=\frac{numberofgoodpens}{totalpens}=\frac{124}{144}$ $=\frac{31}{36}$

Q21 (ii) A lot consists of 144 ball pens, of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that (ii) She will not buy it?

Answer:

Total number of pens = 144

Total number of defective pens = 20

She will buy it if the pen is good.

Therefore, the probability that she will not buy = the probability that the pen is defective =

$P(gettingadefectivepen)=\frac{no.ofdefectivepens}{totalpens}=\frac{20}{144}$ $=\frac{5}{36}$

Q22 (i) Refer to Example 13. (i) Complete the following table:

Answer:

The table becomes:

Q 22 (ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability of 1/11. Do you agree with this argument? Justify your answer.

Answer:

A student argues that there are 11 possible outcomes: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Hence, each of them has a probability of 1/11. We do not agree with this argument because there are a different number of possible outcomes for each sum. We can see that each sum has a different probability.

Q23 A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result, i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.

Answer:

The possible outcomes when a coin is tossed 3 times: {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Therefore, the number of total possible outcomes = 8

For Hanif to win, all the tosses should give the same result, so there are only two favourable outcomes: {HHH, TTT}

Therefore, the number of favourable outcomes = 2

Thus, $P(Hanifwillwin)=\frac{favourableoutcomes}{totaloutcomes}=\frac{2}{8}$ $=\frac{1}{4}$

And, $P(Hanifwilllose)=1-\frac{1}{4}$ $=\frac{3}{4}$

Hence, the probability that Hanif will lose is $\frac{3}{4}$

Q24 (i) A die is thrown twice. What is the probability that 5 will not come up either time?

Answer:

When a die is thrown twice, the possible outcomes = $\{(x,y):x,y\in \{1,2,3,4,5,6\}\}$

Total number of possible outcomes = 6 $\times$ 6 = 36

The outcomes when 5 comes up are either on them =

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

Therefore, the number of such favourable outcomes = 11

Thus, $P(5willcomeupeithertime)=\frac{11}{36}$

Therefore $P(5willnotcomeupeithertime)=1-\frac{11}{36}=\frac{25}{36}$

Hence, the probability that 5 will not come either time is $\frac{25}{36}$

24 (ii) A die is thrown twice. What is the probability that 5 will come up at least once?

Answer:

When a die is thrown twice, the possible outcomes = $\{(x,y):x,y\in \{1,2,3,4,5,6\}\}$

Total number of possible outcomes = 6 $times$ 6 = 36

The outcomes when 5 comes up at least once =

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

Number of such favourable outcomes = 11

Therefore $P(5comesupatleastonce)=\frac{11}{36}$

Hence, the probability that 5 comes at least once is $\frac{11}{36}$

Q25 (i) Which of the following arguments are correct and which are not correct? Give reasons for your answer. (i) If two coins are tossed simultaneously, there are three possible outcomes—two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1/3

Answer:

The possible outcomes when two coins are tossed = {HH, HT, TH, TT}

Therefore, the total number of possible outcomes = 4

Thus $P(gettingtwoheads)=\frac{favourableoutcomes}{totaloutcomes}=\frac{1}{4}$

Hence, the given statement is not correct. This is because one of each can occur in two different ways. Hence, the mentioned events are not equally likely.

Q25 (ii) Which of the following arguments are correct and which are not correct? Give reasons for your answer. (ii) If a die is thrown, there are two possible outcomes—an odd number or an even number. Therefore, the probability of getting an odd number is 1/2

Answer:

The possible outcomes when a die is thrown are {1,2,3,4,5,6}

Total number of possible outcomes = 6

Number of odd numbers, {1,3,5} = 3

And, the number of even numbers {2,4,6} = 3

Hence, both these events are equally likely

$\therefore P(gettinganodd)=\frac{favourableoutcomes}{totaloutcomes}=\frac{3}{6}=\frac{1}{2}$