PROBABILITY

EXERCISE 14.1

Q1. Complete the following statements:

(i) Probability of an event E + Probability of the event ‘not E’ = 1

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

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

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

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

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Q2: 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.

Answer: The outcomes are not equally likely, because the chances of the car starting or not starting depend on various factors such as the condition of the car, fuel, and battery.

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

Answer: The outcomes are not equally likely, because the player’s skill, distance, and practice affect the chances of making or missing the shot.

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

Answer: The outcomes are equally likely, because there are only two possible outcomes—right or wrong—and each has an equal chance of occurring.

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

Answer: The outcomes are almost equally likely, because the probability of having a boy or a girl is nearly the same.

Final answer: (iii) and (iv) are equally likely outcomes.

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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:

Tossing a coin is considered to be a fair way of deciding because it has two equally likely outcomes — heads or tails. Each team has an equal chance of winning the toss, and the result depends purely on chance, not on any team’s skill or influence.

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

(A) 2/3  (B) –1.5  (C) 15%  (D) 0.7

Answer:

(B) –1.5

Explanation:

The probability of an event always lies between 0 and 1. Since –1.5 is less than 0, it cannot be a valid probability.

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

Answer:

P(not E) = 1 – P(E)

    = 1 – 0.05

    = 0.95

Therefore, the probability of ‘not E’ is 0.95.

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Q6: 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?

Answer:

Since all the candies in the bag are lemon flavoured:

(i) Probability of taking out an orange flavoured candy = 0

(ii) Probability of taking out a lemon flavoured candy = 1

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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:

Let the event E = “2 students not having the same birthday.”

Then, the event not E = “2 students having the same birthday.”

So,

P(not E) = 1 – P(E)

    = 1 – 0.992

    = 0.008

Therefore, the probability that the 2 students have the same birthday is 0.008.

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Q8: 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?

Answer:

Total number of balls = 3 + 5 = 8

(i) Probability of drawing a red ball = Number of red balls / Total balls

                    = 3 / 8

(ii) Probability of drawing a ball that is not red = 1 – P(red)

                    = 1 – 3/8

                    = 5/8

Therefore,

(i) P(red) = 3/8

(ii) P(not red) = 5/8

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Q9: 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?

Answer:

Total number of marbles = 5 + 8 + 4 = 17

(i) Probability of red marble = Number of red marbles / Total marbles

               = 5 / 17

(ii) Probability of white marble = Number of white marbles / Total marbles

                 = 8 / 17

(iii) Probability of not green marble = 1 – P(green)

                    = 1 – (4 / 17)

                    = 13 / 17

Therefore,

(i) P(red) = 5/17

(ii) P(white) = 8/17

(iii) P(not green) = 13/17

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Q10: A piggy bank contains hundred 50p coins, fifty ₹1 coins, twenty ₹2 coins and ten ₹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 50p coin? (ii) will not be a ₹5 coin?

Answer:

Total number of coins = 100 + 50 + 20 + 10 = 180

(i) Probability that the coin will be a 50p coin = Number of 50p coins / Total coins

                      = 100 / 180

                      = 5 / 9

(ii) Probability that the coin will not be a ₹5 coin = 1 – P(₹5 coin)

                        = 1 – (10 / 180)

                        = 170 / 180

                        = 17 / 18

Therefore,

(i) P(50p coin) = 5/9

(ii) P(not ₹5 coin) = 17/18

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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. 14.4 (refer book)). What is the probability that the fish taken out is a male fish?

Answer:

Total number of fish = 5 + 8 = 13

Number of male fish = 5

Probability of taking out a male fish = Number of male fish / Total fish

                    = 5 / 13

Therefore, the probability that the fish taken out is a male fish = 5/13.

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Q12: 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. 14.5), 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?

Answer:

Total possible outcomes = 8 (numbers 1 to 8)

(i) Probability that it points at 8 = 1/8

(ii) Odd numbers = 1, 3, 5, 7 → 4 odd numbers

  Probability = 4/8 = 1/2

(iii) Numbers greater than 2 = 3, 4, 5, 6, 7, 8 → 6 numbers

  Probability = 6/8 = 3/4

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

  Probability = 8/8 = 1

Therefore,

(i) P(8) = 1/8

(ii) P(odd number) = 1/2

(iii) P(number > 2) = 3/4

(iv) P(number < 9) = 1

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Q13: 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

Answer:

Total possible outcomes when a die is thrown = 6

(Numbers on a die: 1, 2, 3, 4, 5, 6)

(i) Prime numbers = 2, 3, 5 → 3 outcomes

  Probability = 3/6 = 1/2

(ii) Numbers lying between 2 and 6 = 3, 4, 5 → 3 outcomes

  Probability = 3/6 = 1/2

(iii) Odd numbers = 1, 3, 5 → 3 outcomes

  Probability = 3/6 = 1/2

Therefore,

(i) P(prime number) = 1/2

(ii) P(number between 2 and 6) = 1/2

(iii) P(odd number) = 1/2

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Q14: 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

Answer:

Total number of cards in a deck = 52

(i) King of red colour

There are 2 red kings (King of hearts and King of diamonds).

Probability = 2 / 52 = 1 / 26

(ii) A face card

Each suit has 3 face cards (Jack, Queen, King).

Total face cards = 3 × 4 = 12

Probability = 12 / 52 = 3 / 13

(iii) A red face card

There are 2 red suits (hearts and diamonds), each having 3 face cards.

Red face cards = 3 × 2 = 6

Probability = 6 / 52 = 3 / 26

(iv) The jack of hearts

Only one such card exists.

Probability = 1 / 52

(v) A spade

There are 13 spade cards in a deck.

Probability = 13 / 52 = 1 / 4

(vi) The queen of diamonds

Only one such card exists.

Probability = 1 / 52

Therefore,

(i) P(King of red colour) = 1/26

(ii) P(Face card) = 3/13

(iii) P(Red face card) = 3/26

(iv) P(Jack of hearts) = 1/52

(v) P(Spade) = 1/4

(vi) P(Queen of diamonds) = 1/52

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Q15: Five cards — the ten, jack, queen, king and ace of diamonds — are well-shuffled with their faces 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?

Answer:

Total number of cards = 5

Cards: 10, Jack, Queen, King, Ace

(i) Probability that the card is a queen = 1 / 5

(ii) After the queen is drawn and put aside, number of cards left = 4

(a) Probability that the next card is an ace = 1 / 4

(b) Probability that the next card is a queen = 0 (since the queen has already been removed)

Therefore,

(i) P(Queen) = 1/5

(ii) (a) P(Ace) = 1/4 (b) P(Queen) = 0

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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 taken out is a good one.

Answer:

Total number of pens = 12 + 132 = 144

Number of good pens = 132

Probability that the pen taken out is a good one = Number of good pens / Total pens

                         = 132 / 144

                         = 11 / 12

Therefore, the probability that the pen taken out is a good one = 11/12.

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Q17: (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?

(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?

Answer:

Total number of bulbs = 20

Number of defective bulbs = 4

Number of good (non-defective) bulbs = 20 – 4 = 16

(i) Probability that the bulb drawn is defective = Number of defective bulbs / Total bulbs

                    = 4 / 20

                    = 1 / 5

(ii) Since the bulb drawn first is not defective and not replaced,

Remaining bulbs = 19

Remaining good bulbs = 16 – 1 = 15

Probability that the second bulb is not defective = 15 / 19

Therefore,

(i) P(defective bulb) = 1/5

(ii) P(not defective bulb in second draw) = 15/19

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Q18: 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.

Answer:

Total number of discs = 90

(i) Two-digit numbers:

Two-digit numbers are from 10 to 90 → Total = 81 numbers (10, 11, 12, …, 90)

Probability = 81 / 90 = 9 / 10

(ii) Perfect square numbers between 1 and 90:

Perfect squares = 1, 4, 9, 16, 25, 36, 49, 64, 81 → Total = 9 numbers

Probability = 9 / 90 = 1 / 10

(iii) Numbers divisible by 5:

Multiples of 5 between 1 and 90 = 5, 10, 15, …, 90

Number of such multiples = 90 ÷ 5 = 18

Probability = 18 / 90 = 1 / 5

Therefore,

(i) P(two-digit number) = 9/10

(ii) P(perfect square number) = 1/10

(iii) P(number divisible by 5) = 1/5

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

A B C D E A

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

(i) A? (ii) D?

Answer:

Total possible outcomes = 6

Letters on the die: A, B, C, D, E, A

(i) Number of faces showing A = 2

 Probability of getting A = 2 / 6 = 1 / 3

(ii) Number of faces showing D = 1

 Probability of getting D = 1 / 6

Therefore,

(i) P(A) = 1/3

(ii) P(D) = 1/6

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Q20: Suppose you drop a die at random on the rectangular region shown in Fig. 14.6(refer book). What is the probability that it will land inside the circle with diameter 1 m?

Answer:

Length of rectangle = 3 m

Breadth of rectangle = 2 m

Area of rectangle = 3 × 2 = 6 m²

Diameter of circle = 1 m

Radius of circle = 1/2 m

Area of circle = πr² = π × (1/2)² = π/4 m²

Probability that the die will land inside the circle =

Area of circle / Area of rectangle

= (π/4) ÷ 6

= π / 24

Therefore, the probability that the die will land inside the circle = π/24.

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Q21: 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?

Solution:

Total number of pens = 144

Number of defective pens = 20

Number of good pens = 144 – 20 = 124

(i) Probability that she will buy it

= Number of good pens / Total number of pens

= 124 / 144

= 31 / 36

(ii) Probability that she will not buy it

= Number of defective pens / Total number of pens

= 20 / 144

= 5 / 36

Answer:

(i) Probability that she will buy it = 31/36

(ii) Probability that she will not buy it = 5/36

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Q22: (i) Complete the following table:

Evnt: ‘Sum 2 dice’ 2 3 4 5 6 7 8 9 10 11 12

Probability 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36

(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 1/11.’

Do you agree with this argument? Justify your answer.

Answer:

No, I do not agree with the student’s argument.

Although there are 11 possible sums when two dice are thrown, the outcomes are not equally likely.

Each sum can be obtained in a different number of ways: for example, sum 7 can occur in 6 ways (1+6, 2+5, 3+4, 4+3, 5+2, 6+1), while sum 2 can occur in only 1 way (1+1).

Since each pair of dice faces has an equal chance of appearing (1 out of 36 possible pairs), the probability of each sum depends on how many pairs make that sum.

Therefore, the probabilities are not equal to 1/11, but vary as shown in the completed table above.

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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.

Solution:

When a coin is tossed 3 times,

Total number of possible outcomes = 2 × 2 × 2 = 8

The possible outcomes are:

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

Hanif wins if he gets either all heads or all tails.

Favourable outcomes for winning = HHH, TTT

Number of favourable outcomes = 2

Probability that Hanif wins = 2 / 8 = 1 / 4

Therefore,

Probability that Hanif loses = 1 – Probability that he wins

= 1 – 1/4

= 3/4

Answer:

Probability that Hanif will lose the game = 3/4

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Q24: 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.]

Solution:

When a die is thrown twice,

Total number of possible outcomes = 6 × 6 = 36

(i) Probability that 5 will not come up either time

For one throw, probability that 5 does not appear = 5/6

For two throws,

Probability that 5 does not appear on both throws = (5/6) × (5/6) = 25/36

So, probability that 5 will not come up either time = 25/36

(ii) Probability that 5 will come up at least once

= 1 – Probability that 5 will not come up either time

= 1 – 25/36

= 11/36

Answer:

(i) Probability that 5 will not come up either time = 25/36

(ii) Probability that 5 will come up at least once = 11/36

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Q25: 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.

(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.

Solution:

(i) Not correct

When two coins are tossed simultaneously, the possible outcomes are:

HH, HT, TH, TT

Total number of outcomes = 4

Favourable outcomes for:

• Two heads = HH (1 outcome)

• Two tails = TT (1 outcome)

• One of each = HT, TH (2 outcomes)

Hence, probabilities are:

• P(two heads) = 1/4

• P(two tails) = 1/4

• P(one of each) = 2/4 = 1/2

Therefore, the outcomes do not have equal probabilities, and the argument is not correct.

(ii) Correct

When a die is thrown, the possible outcomes are:

1, 2, 3, 4, 5, 6

Odd numbers: 1, 3, 5 → 3 outcomes

Even numbers: 2, 4, 6 → 3 outcomes

Probability of getting an odd number = 3/6 = 1/2

Hence, the argument is correct.

Answer:

(i) Not correct

(ii) Correct

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