Q.1 Complete the statements:

(i) Probability of event E + Probability of event ‘not E’ = _______

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

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

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

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

(i) Probability of event E + Probability of event ‘not E’ = _______

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

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

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

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

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(i) 1

(ii) 0, impossible event

(iii) 1, sure or certain event

(iv) 1

(v) 0, 1

Q.2 Which of the following experiments have equally likely outcomes? Explain.

(i) A driver attempts to start a car. The car starts or does not start.

(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.

(iii) A trial is made to answer a true-false question. The answer is right or wrong.

(iv) A baby is born. It is a boy or a girl.

(i) A driver attempts to start a car. The car starts or does not start.

(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.

(iii) A trial is made to answer a true-false question. The answer is right or wrong.

(iv) A baby is born. It is a boy or a girl.

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(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 wrong. 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 two possible outcome — 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.

Q.3 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?

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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.

Q.4 Which of the following cannot be the probability of an event?

(A) \( \frac{2}{3} \) (B) \( - 1.5 \)

(C) \( 15 % \) (D) \( 0.7 \)

(A) \( \frac{2}{3} \) (B) \( - 1.5 \)

(C) \( 15 % \) (D) \( 0.7 \)

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∵ The probability of an event E is a number P(E) such that

\( 0 \ \le \ P(E) \ \le \ 1 \)

∴ -1.5 cannot be the probability of an event.

∴ (B) is the correct answer.

Q.5 IF P(F) = 0.05, what is the probability of 'not E' ?

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

=> \( P (not \ E) \ = \ 1 \ - \ P(E) \ = \ 1 \ - \ 0.05 \ = \ 0.95 \)

Q.6 A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out

(i) an orange flavoured candy?

(ii) a lemon flavoured candy?

(i) an orange flavoured candy?

(ii) a lemon flavoured candy?

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(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.

Q.7 It is given that in a group of 3 students, the probability of 2 student not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?

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Let E be the event of having the same birthday

i.e, \( P(E) \ = \ 0.992 \)

But \( P(E) \ + \ P( \overline{E} ) \ = \ 1 \)

\( P( \overline{E}) \ = \ 1 \ - \ P(E) \)

=> \( 1 \ - \ 0.992 \ = \ 0.008 \)

∴ The probability that the 2 students have the same birthday is 0.008.

Q.8 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?

(ii) not red?

(i) red?

(ii) not red?

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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 ) \( = \ \frac{3}{8} \)

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

∴ , Favourable number of elementary events = 5

Hence, P(not getting a red ball) \( = \ \frac{5}{8} \)

Q.9 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?

(ii) white?

(iii) not green?

(i) red?

(ii) white?

(iii) not green?

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Total number of marbles in the box = 5 (Red) + 8 (White) + 4 (Green) = 17

∴ Total number of elementary events = 17

(i) There are 5 red marbles in the box

∴ Favourable number of elementary events = 5

Hence, P (getting a red marble) \( = \ \frac{5}{17} \)

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

∴ Favourable number of elementary events = 8

Hence, P (getting a white marble) \( = \ \frac{8}{17} \)

(iii) There are 5 + 8 = 13 marbles which are not green in the box.

∴ Favourable number of elementary events = 13

Hence, P (not getting a green marble) \( = \ \frac{13}{17} \)

Q.10 A piggy bank contains hundred 50 p coins, fifty Re 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

(i) will be a 50 p coin?

(ii) will not be a Rs 5 coin?

(i) will be a 50 p coin?

(ii) will not be a Rs 5 coin?

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

∴ Total number of elementary events = 180

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

∴ Favourable number of elementary events = 100

Hence, P(falling out of a 50p coin) \( = \ \frac{100}{180} \ = \ \frac{5}{9} \)

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

∴ Favourable number of elementary events = 170

Hence, P(falling out of a coin other than Rs 5 coin)

\( \frac{170}{180} \ = \ \frac{17}{18} \)

Q.11 Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fishes and 8 females fishes (see fig.). What is the probability that the fish taken out is a male fish?

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The total number of fish in the tank = 5 + 8 = 13

Total number of male fish = 5

\(P(E) \ = \ \frac{Number \ of \ favourable \ outcomes}{Total \ number \ of \ outcomes} \)

P (male fish) \(= \ \frac{5}{13} \ = \ 0.38 \)

12. 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

(i) 8?

(ii) an odd number?

(iii) a number greater than 2?

(iv) a number less than 9?

(i) 8?

(ii) an odd number?

(iii) a number greater than 2?

(iv) a number less than 9?

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Total number of possible outcomes = 8

\( P(E) \ = \ \frac{Number \ of \ favourable \ outcomes}{Total \ number \ of \ outcomes} \)

(i) Total number of favourable events (i.e. 8) = 1

P(pointing at 8) \( = \ \frac{1}{8} \ = \ 0.125 \)

(ii) Total number of odd numbers = 4 (1, 3, 5 and 7)

P(pointing at an odd number) \( = \ \frac{4}{8} \ = \ 0.5 \)

(iii) Total numbers greater than 2 = 6 (3, 4, 5, 6, 7 and 8)

P(pointing at a number greater than 4) \( = \ \frac{6}{8} \ = \ 0.75 \)

(iv) Total numbers less than 9 = 8 (1, 2, 3, 4, 5, 6, 7, and 8)

P(pointing at a number less than 9) \( = \ \frac{8}{8} \ = \ 1 \)

13. A die is thrown once. Find the probability of getting

(i) a prime number

(ii) a number lying between 2 and 6

(iii) an odd number.

(i) a prime number

(ii) a number lying between 2 and 6

(iii) an odd number.

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Total possible events when a dice is thrown = 6 (1, 2, 3, 4, 5, and 6)

\( P(E) \ = \ \frac{Number \ of \ favourable \ outcomes}{Total \ number \ of \ outcomes} \)

(i) Total number of prime numbers = 3 (2, 3 and 5)

P (getting a prime number) \(= \ \frac{3}{6} \ = \ 0.5 \)

(ii) Total numbers lying between 2 and 6 = 3 (3, 4 and 5)

P (getting a number between 2 and 6) \( = \ \frac{3}{6} \ = \ 0.5 \)

(iii) Total number of odd numbers = 3 (1, 3 and 5)

P (getting an odd number) \( = \ \frac{3}{6} \ = \ 0.5 \)

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

(i) a king of red colour

(ii) a face card

(iii) a red face card

(iv) the jack of hearts

(v) a spade

(vi) the queen of diamonds

(i) a king of red colour

(ii) a face card

(iii) a red face card

(iv) the jack of hearts

(v) a spade

(vi) the queen of diamonds

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Total number of possible outcomes = 52

\( P(E) \ = \ \frac{Number \ of \ favourable \ outcomes}{Total \ number \ of \ outcomes} \)

(i) Total numbers of king of red colour = 2

P (getting a king of red colour) \(= \ \frac{2}{52} \ = \ \frac{1}{26} \ = \ 0.038 \)

(ii) Total numbers of face cards = 12

P (getting a face card) \(= \ \frac{12}{52} \ = \ \frac{3}{13} \ = \ 0.23 \)

(iii) Total numbers of red face cards = 6

P (getting a king of red colour) \(= \ \frac{6}{52} \ = \ \frac{3}{26} \ = \ 0.11 \)

(iv) Total numbers of jack of hearts = 1

P (getting a king of red colour) \(= \ \frac{1}{52} \ = \ 0.019 \)

(v) Total numbers of king of spade = 13

P (getting a king of red colour) \(= \ \frac{13}{52} \ = \ \frac{1}{4} \ = \ 0.25 \)

(vi) Total numbers of queen of diamonds = 1

P (getting a king of red colour) \(= \ \frac{1}{52} \ = \ 0.019 \)

15. 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.

(i) What is the probability that the card is the queen?

(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?

(i) What is the probability that the card is the queen?

(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?

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Total numbers of cards = 5

\( P(E) \ = \ \frac{Number \ of \ favourable \ outcomes}{Total \ number \ of \ outcomes} \)

(i) Numbers of queen = 1

P (picking a queen) \( = \ \frac{1}{5} \ = \ 0.2 \)

(ii) If the queen is drawn and put aside, the total numbers of cards left is (5-1) = 4

(a) Total numbers of ace = 1

P (picking an ace) \( = \ \frac{1}{4} \ = \ 0.25 \)

(b) Total numbers of queen = 0

P (picking a queen) \( = \ \frac{0}{4} \ = \ 0 \)

16. 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 taken out is a good one.

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Numbers of pens = Numbers of defective pens + Numbers of good pens

∴ Total number of pens = 132 + 12 = 144 pens

\( P(E) \ = \ \frac{Number \ of \ favourable \ outcomes}{Total \ number \ of \ outcomes} \)

P(picking a good pen) \( = \ \frac{132}{144} \ = \ \frac{11}{12} \ = \ 0.916 \)

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

(ii) Suppose the bulb 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?

(ii) Suppose the bulb 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?

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(i) Numbers of defective bulbs = 4

The total numbers of bulbs = 20

\( P(E) \ = \ \frac{Number \ of \ favourable \ outcomes}{Total \ number \ of \ outcomes} \)

∴ Probability of getting a defective bulb = P (defective bulb) \( = \ \frac{4}{20} \ = \ \frac{1}{5} \ = \ 0.2 \)

(ii) Since 1 non-defective bulb is drawn, then the total numbers of bulbs left are 19

So, the total numbers of events (or outcomes) = 19

Numbers of defective bulbs = 19 - 4 = 15

So, the probability that the bulb is not defective \( = \ \frac{15}{19} \ = \ 0.789 \)

18. 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

(i) a two-digit number

(ii) a perfect square number

(iii) a number divisible by 5.

(i) a two-digit number

(ii) a perfect square number

(iii) a number divisible by 5.

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The total numbers of discs = 50

\( P(E) \ = \ \frac{Number \ of \ favourable \ outcomes}{Total \ number \ of \ outcomes} \)

(i) Total number of discs having two digit numbers = 81

(∵ 1 to 9 are single digit numbers and so, total 2 digit numbers are 90 - 9 = 81)

P (bearing a two-digit number) \( = \ \frac{81}{90} \ = \ \frac{9}{10} \ = \ 0.9 \)

(ii) Total number of perfect square numbers = 9 (1, 4, 9, 16, 25, 36, 49, 64 and 81)

P (getting a perfect square number) \( =\ \frac{9}{90} \ = \ \frac{1}{10} \ = \ 0.1 \)

(iii) Total numbers which are divisible by 5 = 18 (5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85 and 90)

P (getting a number divisible by 5) \( = \ \frac{18}{90} \ = \ \frac{1}{5} \ = \ 0.2 \)

19. 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?

(ii) D?

The die is thrown once. What is the probability of getting

(i) A?

(ii) D?

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The total number of possible outcomes (or events) = 6

\( P(E) \ = \ \frac{Number \ of \ favourable \ outcomes}{Total \ number \ of \ outcomes} \)

(i) The total number of faces having A on it = 2

P (getting A) \( = \ \frac{2}{6} \ = \ \frac{1}{3} \ = \ 0.33 \)

(ii) The total number of faces having D on it = 1

P (getting D) \( = \ \frac{1}{6} \ = \ 0.166 \)

20. 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 diameter 1m?

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First, calculate the area of the rectangle and the area of the circle. Here, the area of the rectangle is the possible outcome and the area of the circle will be the favourable outcome.

So, the area of the rectangle = (3 × 2) m^{2} = 6 m^{2}

and,

The area of the circle \( = \ \pi r^2 \ = \ \pi ( \frac{1}{2})^2 \ = \ \frac{ \pi}{4} \ = \ 0.78 \)

∴ The probability that die will land inside the circle \( = \ \frac{ \frac{ \pi}{4}}{6} \ = \ \frac{\pi}{24} \ = \ 0.13 \)

21. 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?

(ii) She will not buy it?

(i) She will buy it?

(ii) She will not buy it?

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The total numbers of outcomes i.e. pens = 144

Given, numbers of defective pens = 20

∴ The numbers of non defective pens = 144 - 20 = 124

\( P(E) \ = \ \frac{Number \ of \ favourable \ outcomes}{Total \ number \ of \ outcomes} \)

(i) Total numbers events in which she will buy them = 124

So, P (buying) \( = \ \frac{124}{144} \ = \ \frac{31}{36} \ = \ 0.86 \)

(ii) Total numbers events in which she will not buy them = 20

So, P (not buying) \(= \ \frac{20}{144} \ = \ \frac{5}{36} \ = \ 0.138 \)

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

(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 \( \frac{1}{11} \). Do you agree with this argument? Justify your Solution:.

Event:
Sum on 2 dice |
2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |

Probability | \( \frac{1}{36} \) | \( \frac{5}{36} \) | \( \frac{1}{36} \) |

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If 2 dices are thrown, the possible events 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)

So, the total numbers of events: 6 × 6 = 36

(i) It is given that to get the sum as 2, the probability is \( \frac{1}{36} \) as the only possible outcomes = (1,1)

For getting the sum as 3, the possible events (or outcomes) = E (sum 3) = (1,2) and (2,1)

So, P(sum 3) \( = \ \frac{2}{36} \)

Similarly,

E (sum 4) = (1,3), (3,1), and (2,2)

So, P (sum 4) \( = \ \frac{3}{36} \)

E (sum 5) = (1,4), (4,1), (2,3), and (3,2)

So, P (sum 5) \(= \ \frac{4}{36} \)

E (sum 6) = (1,5), (5,1), (2,4), (4,2), and (3,3)

So, P (sum 6) \( = \ \frac{5}{36} \)

E (sum 7) = (1,6), (6,1), (5,2), (2,5), (4,3), and (3,4)

So, P (sum 7) \(= \ \frac{6}{36} \)

E (sum 8) = (2,6), (6,2), (3,5), (5,3), and (4,4)

So, P (sum 8) \(= \ \frac{5}{36} \)

E (sum 9) = (3,6), (6,3), (4,5), and (5,4)

So, P (sum 9) \(= \ \frac{4}{36} \)

E (sum 10) = (4,6), (6,4), and (5,5)

So, P (sum 10) \( = \ \frac{3}{36} \)

E (sum 11) = (5,6), and (6,5)

So, P (sum 11) \( = \ \frac{2}{36} \)

E (sum 12) = (6,6)

So, P (sum 12) \( = \ \frac{1}{36} \)

So, the table will be as:

Event:
Sum on 2 dice |
2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |

Probability | \( \frac{1}{36} \) | \( \frac{2}{36} \) | \( \frac{3}{36} \) | \( \frac{4}{36} \) | \( \frac{5}{36} \) | \( \frac{6}{36} \) | \( \frac{5}{36} \) | \( \frac{4}{36} \) | \( \frac{3}{36} \) | \( \frac{2}{36} \) | \( \frac{1}{36} \) |

(ii) The argument is not correct as it is already justified in (i) that the number of all possible outcomes is 36 and not 11.

23. 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.

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The total number of outcomes = 8 (HHH, HHT, HTH, THH, TTH, HTT, THT, TTT)

Total outcomes in which Hanif will lose the game = 6 (HHT, HTH, THH, TTH, HTT, THT)

P (losing the game) \( = \ \frac{6}{8} \ = \ \frac{3}{4} \ = \ 0.75 \)

24. A die is thrown twice. What is the probability that

(i) 5 will not come up either time?

(ii) 5 will come up at least once?

[Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]

(i) 5 will not come up either time?

(ii) 5 will come up at least once?

[Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]

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Outcomes 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)

So, the total number of outcome = 6 × 6 = 36

(i) **Method 1:**

Consider the following events.

A = 5 comes in first throw,

B = 5 comes in second throw

P(A) \( = \ \frac{6}{36} \),

P(B) \( = \ \frac{6}{36} \) and

P(not B) \( = \ \frac{5}{6} \)

So, P(not A) \( = \ 1 \ - \ \frac{6}{36} \ = \ \frac{5}{6} \)

The required probability \(= \ \frac{5}{6} \ × \ \frac{5}{6} \ = \ \frac{25}{36} \)

**Method 2:**

Let E be the event in which 5 does not come up either time.

So, the favourable outcomes are [36 - (5 + 6)] = 25

∴ P(E) \( = \ \frac{25}{36} \)

(ii) Number of events when 5 comes at least once = 5 + 6 = 11

∴ The required probability \( = \ \frac{11}{36} \)

25. Which of the following arguments are correct and which are not correct? Give reasons for your Solution:

(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 \( \frac{1}{3} \)

(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 \( \frac{1}{2} \)

(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 \( \frac{1}{3} \)

(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 \( \frac{1}{2} \)

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(i) All the possible events are (H,H); (H,T); (T,H) and (T,T)

So, P (getting two heads) \( = \ \frac{1}{4} \)

and, P (getting one of the each) \( = \ \frac{2}{4} \ = \ \frac{1}{2} \)

∴ This statement is incorrect.

(ii) Since the two outcomes are equally likely, this statement is correct.

1. Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on

(i) the same day?

(ii) consecutive days?

(iii) different days?

(i) the same day?

(ii) consecutive days?

(iii) different days?

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Since there are 5 days and both can go to the shop in 5 ways each so,

The total number of possible outcomes = 5 × 5 = 25

(i) The number of favourable events = 5 (Tue., Tue.), (Wed., Wed.), (Thu., Thu.), (Fri., Fri.), (Sat., Sat.)

So, P (both visiting on the same day) \(= \ \frac{5}{25} \ = \ \frac{1}{5} \)

(ii) The number of favourable events = 8 (Tue., Wed.), (Wed., Thu.), (Thu., Fri.), (Fri., Sat.), (Sat., Fri.), (Fri., Thu.), (Thu., Wed.), and (Wed., Tue.)

So, P(both visiting on the consecutive days) \( = \ \frac{8}{25} \)

(iii) P (both visiting on the different days) = 1 - P (both visiting on the same day)

So, P (both visiting on the different days) \( = \ 1 \ - \ \frac{1}{5} \ = \ \frac{4}{5} \)

2. A die is numbered in such a way that its faces show the numbers 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws:

What is the probability that the total score is

(i) even?

(ii) 6?

(iii) at least 6?

What is the probability that the total score is

(i) even?

(ii) 6?

(iii) at least 6?

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The table will be as follows:

+ | 1 | 2 | 2 | 3 | 3 | 6 |

1 | 2 | 3 | 3 | 4 | 4 | 7 |

2 | 3 | 4 | 4 | 5 | 5 | 8 |

2 | 3 | 4 | 4 | 5 | 5 | 8 |

3 | 4 | 5 | 5 | 6 | 6 | 9 |

3 | 4 | 5 | 5 | 6 | 6 | 9 |

6 | 7 | 8 | 8 | 9 | 9 | 12 |

So, the total number of outcome = 6 × 6 = 36

(i) E (Even) = 18

P (Even) \( = \ \frac{18}{36} \ = \ \frac{1}{2} \)

(ii) E (sum is 6) = 4

P (sum is 6) \( = \ \frac{4}{36} \ = \ \frac{1}{9} \)

(iii) E (sum is atleast 6) = 15

P (sum is atleast 6) \( = \ \frac{15}{36} \ = \ \frac{5}{12} \)

3. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.

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It is given that the total number of red balls = 5

Let the total number of blue balls = x

So, the total no. of balls = x + 5

P(E) = (Number of favourable outcomes/ Total number of outcomes)

P (drawing a blue ball) \( = \ \frac{x}{x \ + \ 5} \) (i)

Similarly,

P (drawing a red ball) \( = \ \frac{5}{x \ + \ 5} \) (ii)

From equation (i) and (ii)

x = 10

So, the total number of blue balls = 10

4. A box contains 12 balls out of which x are black. If one ball is drawn at random from the box what is the probability that it will be a black ball ? If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.

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Total number of black balls = x

Total number of balls = 12

\( P(E) \ = \ \frac{Number \ of \ favourable \ outcomes}{Total \ number \ of \ outcomes} \)

P (getting black balls) \(= \ \frac{x}{12} \) (i)

Now, when 6 more black balls are added,

Total balls become = 18

∴ Total number of black balls = x + 6

Now, P (getting black balls) \( = \ \frac{x \ + \ 6}{18} \) (ii)

Solving equation (i) and (ii)

x = 3

5. A jar contains 24 marbles, some are green and other are blue. If a marble is drawn at random from the jar, the probability that it is green is 23. and the number of blue balls in the jar.

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Total marbles = 24

Let the total green marbles = x

So, the total blue marbles = 24 - x

P(getting green marble) \( = \ \frac{x}{24} \)

From the question, \( \frac{x}{24} \ = \ \frac{2}{3} \)

So, the total green marbles = 16

And, the total blue marbles = 24 - x = 8