Haryana State Board HBSE 10th Class Maths Solutions Chapter 15 Probability Ex 15.1 Textbook Exercise Questions and Answers.

## Haryana Board 10th Class Maths Solutions Chapter 15 Probability Ex 15.1

Question 1.

Complete the following statements:

(i) ProbabilIty of an event E + Probability of the 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 elementry events of an experiment is …………………. .

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

Solution:

(i) 1

(ii) 0, impossible event

(iii) 1, sure or certain event

(iv) 1

(v) 0, 1.

Question 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 basket ball. 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 bady is born. It is a boy or a girl.

Solution:

(i) The outcome is not equally, likely because when there is some defect the car does not start.

(ii) The outcome in this situation is not equally likely because the outcome depends on many factors such quality of the gun, training of the player etc.

(iii) The outcome in this situation are equally likely because out come in this trial the answer is right or wrong i.e., one out of the two and both have equal chances to happen.

(iv) The outcome in this situation are equally likely because a baby is bom can be either a boy or girl.

Question 3.

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

Solution:

When we toss a coin, the outcomes head and tail are equally likely. So the result of an individual coin toss is completely unpredictable.

Question 4.

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

(a) \(\frac{2}{3}\)

(b) – 1.5

(c) 15%

(d) 0.7

Solution:

The value of probability of an event cannot be negative or greater than 1.

So, the correct option is (B).

Question 5.

If P (E) = 0.05, what is the probability of “not E” ?

Solution:

We have,

P(E) = 0.05

P (not E) = 1- P(E)

= 1 – 0.05 = 0.95

Hence, P(not E) = 0.95.

Question 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?

Solution:

(i) Let E denote the event to get an orange flavoured candy.

There is no outcome favourable to the event E because all the candies are lemon flavoured.

So, P(E) = 0.

(ii) Let F denote the event to get a lemon flavoured candy.

All the outcomes favourable to event F because all the candies in the bag lemon flavoured.

So P(F) = 1.

Hence, (i) P(E) = 0

(ii) P(F) = 1.

Question 7.

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?

Solution:

Let E denote the event that two students out of three have same birthday.

P(not E) = 0.992 (given)

Then P(E) = 1 – P (not E)

= 1 – 0992

= 0008.

Hence, P(E) = 0.008.

Question 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?

Solution:

Number of all possible out comes = 3 + 5 = 8

(i)Number of red balls = 3

Number of favourable outcomes = 3

∴ P(redball) = \(\frac{3}{8}\)

(ii)Number of not red balls = black balls = 5

Number of favourable outcomes = 5

P(not red ball) = \(\frac{5}{8}\)

Hence, (i) P(red ball) = \(\frac{3}{8}\),

(ii) P (not red ball) = \(\frac{5}{8}\).

Question 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 ?

Solution:

Total number of possible out comes = 5 + 8 + 4 = 17

(i) Number of red marbles = 5

Number of favourable outcomes = 5

P(a red marble) = \(\frac{5}{17}\).

(ii) Number of white marbles = 8

Number of favourable outcomes = 8

P(a white marble) = \(\frac{8}{17}\).

(iii) Number of marbles which are not green = 5 + 8 = 13

Number of favourable outcomes = 13

∴ P(a marble not green) = \(\frac{13}{17}\).

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

(ii) P(a white marble) = \(\frac{8}{17}\)

(iii) P(a marble not green) = \(\frac{13}{17}\).

Question 10.

A piggy bank contains hundred 50 P 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 50 P coin ?

(ii) will not be a ₹ 5 coin?

Solution:

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

Total number of possible outcomes = 180

(i) Number of 50 paise coins = 100

Number of favourable outcomes = 100

∴ P(50 paise coins) = \(\frac{100}{180}=\frac{5}{9}\).

(ii) Number of coins other than 5 coins 100 + 50 + 20 = 170

Number of favourable outcomes = 170

P(will not be ₹ 5 coins) = \(\frac{170}{180}=\frac{17}{18}\)

Hence, (i) P (50 P coins) = \(\frac{5}{9}\)

(ii) P(will not be ₹ 5 coins) = \(\frac{17}{18}\).

Question 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 fish and 8 female fish (see in figure). What is the probability that the fish taken out is a male fish?

Solution:

Total fish in the aquarium = 5 + 8 = 13

Total number of possible outcomes = 13

Number of male fish = 5

Number of favourable outcomes = 5

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

Hence, P(male fish) = \(\frac{5}{13}\).

Question 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 in figure), and these are equally likely outcomes. What is the probability that it will point at [CBSE 2016]

(i) 8 ?

(ii) an odd number ?

(iii) a number greater than 2 ?

(iv) a number less than 9 ?

Solution:

Since arrow can come to rest at any one of the numbers 1, 2, 3, 4, 5, 6, 7, 8.

Total number of possible outcomes = 8

(i) Number of favourable outcomes = 1

∴ P(Arrow point at 8) = \(\frac{1}{8}\).

(ii) Odd numbers are 1, 3, 5, 7

Number of favourable outcomes = 4

∴ P(Arrow points at an odd number) = \(\frac{4}{8}=\frac{1}{2}\)

(iii) The numbers greater than 2 are 3, 4, 5, 6, 7 , 8

Number of favourable outcomes = 6

∴ P(Arrow points a number greater than 2 = \(\frac{6}{8}=\frac{3}{4}\).

(iv) The numbers less than 9 are 1, 2, 3, 4, 5, 6, 7, 8.

Number of favourable outcomes = 8.

∴ P(Arrow points at a number less than 9 = \(\frac{8}{8}\) = 1.

Hence, (i) P (Arrow point at 8) = \(\frac{1}{8}\)

(ii) P(Arrow points at an odd number) = \(\frac{1}{2}\)

(iii) P(Arrow points at a number greater them 2) = \(\frac{3}{4}\)

(iv) P(Arrow points at a number less than 9) = 1.

Question 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.

Solution:

In a single throw of a die, all possible out comes are 1, 2, 3, 4, 5, 6.

Total number of possible outcomes = 6

(i) Let E_{1} be event of getting a prime number. Prime numbers are 2, 3, 5

Number of favourable outcomes = 3

∴ P (getting a prime number) = p(E_{1}) = \(\frac{3}{6}=\frac{1}{2}\).

(ii) Let E_{2} be event of getting a number lying between 2 and 6

The numbers lying between 2 and 6 are 3, 4, 5.

Number of favourable outcomes = 3

P(E_{2}) = \(\frac{3}{6}=\frac{1}{2}\).

(iii) Let E_{3} be event of getting an odd number The odd numbers are 1, 3, 5

Number of favourable outcomes = 3

P(E_{3}) = \(\frac{3}{6}=\frac{1}{2}\).

Hence, (i) P(E_{1}) = \(\frac{1}{2}\)

(ii) P(E_{2}) = \(\frac{1}{2}\)

(iii) P(E_{3}) = \(\frac{1}{2}\).

Question 14.

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

(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.

Solution:

Total number of all possible out comes = 52

(i) Number of king of red colour = 2

Number of favourable out comes = 2

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

(ii) We know that kings, queens and jacks are the face cards

Number of face cards = 12

Number of favourable outcomes = 12

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

(iii) Number of red face cards = 6

Number of favourable outcomes = 6

P(getting a red face card) = \(\frac{6}{52}=\frac{3}{26}\).

(iv) Number of jack of hearts = 1

Number of favourable outcomes = 1

P(getting the jack of hearts) = \(\frac{1}{52}\)

(v) Number of spades = 13

Number of favourable outcomes = 13

P(getting a spade) = \(\frac{13}{52}=\frac{1}{4}\).

(vi) Number of queen of diamonds = 1

Number of favourable outcome = 1

P (getting the queen of diamonds) = \(\frac{1}{52}\)

Hence, (i) P(getting a king of red colour) = \(\frac{1}{26}\)

(ii) P (getting a face card) = \(\frac{3}{13}\)

(iii) PCgetting a red face card) = \(\frac{3}{26}\)

(iv) P(getting the jack of hearts) = \(\frac{1}{52}\)

(v) P(getting a spade) = \(\frac{1}{4}\)

(vi) P(getting the queen of diamonds) = \(\frac{1}{52}\).

Question 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 ?

Solution:

(i) The total number of cards = 5

Total number of possible outcomes = 5

Out of these 5 cards only one is queen

Number of favourable outcomes = 1

∴ P(a queen) = \(\frac{1}{5}\)

(ii) After the queen has been drawn and put a side total number of cards = 5 – 1 = 4

Number of possible outcomes = 4

These four cards include one ace and no queen

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

P (a queen) = 0.

Hence, (i) P(a queen) = \(\frac{1}{5}\)

(ii) (a) P(an ace) = \(\frac{1}{4}\)

(b) P(a queen) = 0.

Question 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.

Solution:

Number of total pens = 12 + 132 = 144

Total number of possible outcomes = 144

Number of good pens = 132

Number of favourable outcomes = 132

P(good pens) = \(\frac{132}{144}=\frac{11}{12}\)

Hence, P (good pens) = \(\frac{11}{12}\).

Question 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 ?

Solution:

(i) Total number of bulbs = 20

Total number of possible outcomes = 20

Number of bulbs defective in this lot = 4

Number of favourable outcomes = 4

P(a defective bulb) = \(\frac{4}{20}=\frac{1}{5}\)

(ii) According to question, the bulb drawn is not defective and it is not replaced.

Total number of bulbs contains the lot = 19

Total number of possible outcomes = 19

Number of good bulbs are = 19 – 4 = 15

Number of favourable outcomes = 15

∴ P(a non defective bulb) = \(\frac{15}{19}\)

Hence, (i) P (a defective bulb) = \(\frac{1}{5}\)

(ii) P(a non defective bulb) = \(\frac{15}{19}\).

Question 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.

Solution:

Total number of possible outcomes = 90

(i) Two digit numbers from 1 to 90 are 10, 11, 12, 13, ……………. 90

Total two digit numbers = 81

Number of favourable outcomes = 81

∴ P(a disc bearing a two digit number) = \(\frac{81}{90}=\frac{9}{10}\).

(ii) Perfect square numbers from 1 to 90 are 1, 4, 9, 16, 25, 36, 49, 64, 81.

Total perfect square numbers = 9

Number of favourable outcomes = 9

∴ P(a disc bearing a perfect square number) = \(\frac{9}{90}=\frac{1}{10}\).

(iii) The numbers divisible by 5 from 1 to 90 are 5, 10, 15, 20, ……………….. 90.

Total numbers divisible by 5 from 1 to 90 = 18

Number of favourable outcomes = 18

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

Hence, (i) P(a disc bearing a two digit number) = \(\frac{9}{10}\)

(ii) P(disc bearing a perfect square number) = \(\frac{1}{10}\)

(iii) P(a disc bearing a number divisible by 5) = \(\frac{1}{5}\).

Question 19.

A child has a die whose six faces shows 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 ?

Solution:

In throwing the die any one of the six faces may come upward.

Total number of possible outcomes = 6

(i) Since there are two faces with letter A .

∴ Number of favourable outcomes = 2

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

(ii) There is one face with letter D

Number of favourable outcome = 1

P(gettings D) = \(\frac{1}{6}\)

Hence, (i) P(getting A) = \(\frac{1}{3}\)

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

Question 20.

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

Solution:

Total area of the rectangular region = l × b

= 3 × 2 = 6 m^{2}

Area of the circular region = πr^{2}

= π × \(\left(\frac{1}{2}\right)^2\) = \(\frac{\pi}{4}\) m^{2}.

P(the die lands inside the circle) = \(\frac{\text { Area of the circular region }}{\text { Total area of rectangular region }}\)

= \(\frac{\frac{\pi}{4}}{6}=\frac{\pi}{24}\)

Hence, P(the die lands inside the circle) = \(\frac{\pi}{24}\).

Question 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 ?

Solution:

Total number of pens = 144

Total number of possible outcomes = 144

(i)Number of defective pens = 20

The number of good pens = 144 – 20 = 124

Number of favourable outcomes = 124

P (Nuri will buy good pen) = \(\frac{124}{144}=\frac{31}{36}\).

(ii) P (Nuri will not buy a good pen) = P

(selecting a defective pen) = \(\frac{20}{144}=\frac{5}{36}\)

Hence, (i) P (Nuri will buy a good pen) = \(\frac{31}{36}\)

(ii) P (Nuri will not buy a good pen) = \(\frac{5}{36}\).

Question 22.

Refer to example 25 of NCERT (i) complete the following table:

(ii) A student argues that ‘there are 11 possible out comes 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.

Solution:

Total number of possible outcomes of the toss of the pair of two dice = 36

(i) (1, 2), (2, 1), favour the event of the getting the sum 3

∴ P(getting the sum 3) = \(\frac{2}{36}\).

Cases (1, 3), (2, 2), (3, 1) favour the event of getting sum 4

∴ P(getting the sum 4) = \(\frac{3}{36}\).

Cases (1, 4), (2, 3), (3, 2), (4, 1) favour the event of getting the sum 5

∴ P(getting the sum 5) = \(\frac{4}{36}\)

Cases (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) favour the event of getting the sum 6

∴ P(getting the sum 6) = \(\frac{5}{36}\)

Cases (1, 6), (2, 6), (3, 4), (4, 3), (6, 2), (6, 1) favour the event of getting the sum 7

∴ P(getting the sum 7) = \(\frac{6}{36}\)

Cases (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) favour the event of getting the sum 8

∴ P(getting the sum 8) = \(\frac{5}{36}\)

Cases (3, 6), (4, 5), (5, 4), (6, 3) favour the event of getting the sum 9

∴ P(getting the sum 9) = \(\frac{4}{36}\)

Cases (4, 6), (5, 5), (6, 4) favour the event of getting the sum 10

∴ P(getting the sum 9) = \(\frac{3}{36}\)

Cases (5, 6), (6, 5), (6, 4) favour the event of getting the sum 11

∴ P(getting the sum 9) = \(\frac{2}{36}\)

(ii) No, as the eleven different outcomes are not equally likely.

Question 23.

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

Solution:

When a coin tossed three times, the possible outcomes are (H, H, H), (H, H, T), (H, T, T), (H, T, H), (T, H, H), (T, H, T), (T, T, H) and (T, T, T)

Total number of possible outcomes = 8

Number of cases when all the tosses give the same results are two i.e., (H, H, H) and (T, T, T).

It is given that Hanif wins when all the tosses give the same results,

Number of cases in which Hanif will lose the game = 8 – 2 = 6

Number of favorable outcomes = 6

∴ P (Hanif will lose the game) = \(\frac{6}{8}=\frac{3}{4}\).

Hence, P (Hanif will lose the game) =\(\frac{3}{4}\).

Question 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.]

Solution:

A die is thrown twice Total number of possible outcomes = 36

(i) Out of these 36 outcomes, 5 comes at least once in the following cases :

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

Total number of these cases = 11

∴ Number of cases when 5 will not come up either time = 36 – 11 = 25

Number of favourable outcomes = 25

∴ P(5 will not come up either time) = \(\frac{25}{36}\).

(ii) Number of cases when 5 comes up at least once = 11

Number of favourable outcomes = 11

∴ P (5 will come up at least once) = \(\frac{11}{36}\)

Hence, (i) P (5 will not come up either time) = \(\frac{25}{36}\)

(ii) P (5 will come up at least once) = \(\frac{11}{36}\).

Question 25.

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 ofeach. 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:

When two coins are tossed.

(i) The possible outcomes are (H, H), (H, T), (T, H), (T, T)

Total number of possible outcomes = 4

P(H, H) = \(\frac{1}{4}\)

P(T, T) = \(\frac{1}{4}\)

P(H, T and T, H) = \(\frac{2}{4}=\frac{1}{2}\)

Hence, argument is incorrect.

(ii) When a die is thrown The possible outcomes are 1, 2, 3, 4, 5, 6

Total number of possible outcomes = 6

We have odd numbers or even numbers

Odd numbers are 1, 3, 5

Number of favourable outcomes = 3

Probability (getting an odd number) = \(\frac{3}{6}=\frac{1}{2}\)

Hence, argument is correct.