Comprehensive NCERT solutions, Q&A of Probability of Class 10 Mathematics. As On <DM>.

Q:

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 elementary events of an experiment is .

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

A:

Probability of an event + probability of an event not E = 1
Probability of an event that cannot happen is 0. Such event is called an impossible event.
The probability of an event that is certain to happen is 1. Such event is sure event.
The sum of probabilities of all the elementary events of an Experiment is 1.
Probability of an event is great than or equal to zero and less than or equal to 1.

Q:

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 (i) will be a 50 p coin ? (ii) will not be Rs 5 coin?

A:

Total no. of coins = 180

No. Of 50 paise coins = 100

No. Of 1-rupee coins = 50

No. Of 2-rupee coin = 20

No. Of 5-rupee coin = 10

(i) Probability of getting 50 paise coin = 100/180 = 5/9

(ii) No. Of not five-rupee coin = 170

Probability of not getting 5-rupee coin

= 170/180 = 17/18

Q:

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?

A:

No. of male fishes = 5

No. Of female fishes = 8

Total no. of fishes = 13

Probability of getting a male fish = 5/13

Q:

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?

A:

(i) Probability of getting number 8 = 1/8

(ii) Total odd numbers on the wheel = 4

Probability of getting an odd number = 4/8 = ½

(iii) Number greater than 2 = 6

Probability of getting no greater than 2 = 6/8 = ¾

(iv) Numbers less than 9 = 8

Probability of getting a no. Less than 9 = 8/8 = 1

Q:

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.

A:

Total no. Of possible outcomes = 6

(i) Prime numbers = 3

Probability of getting a prime no. = 3/6 = ½

(ii) Numbers between 2 and 6 = 3

Probability of getting a no. between 2 and 6

= 3/6 = ½

(iii) Odd numbers = 3

Probability of getting an odd no. = 3/6 = ½

Q:

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

A:

Total no. Of cards = 52

(I) Numbers of king of red color = 2

Probability of getting king of red color =2/52=1/26

(ii) Number of face cards = 12

Probability of getting a face card = 12/52 = 3/13

(iii) Number of red face cards = 6

Probability of getting a face card = 6/52 = 3/26

(iv) Number of jack of hearts = 1

Probability of getting a jack of heart = 1/52

(v) Number of a spade = 13

Probability of getting a spade = 13/52 = ¼

(vi) Number of queens of diamond = 1

Probability of getting a queen of diamond = 1/52

Q:

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?

A:

Total no. Of cards = 5

(i) Number of cards of queen = 1

Probability of getting a queen = 1/5

(ii) Now, keeping the queen aside only four cards are left

So,

Total no. of outcome = 4

(a) Number of ace cards = 1

Probability of getting an ace = ¼

(b) Number of queen cards = 0

Probability of getting a card of queen = 0/4 = 0

Q:

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.

A:

Total no. Of pens = 144

Number of defective pens = 12

Number of good pens = 132

Probability of getting a good pen = 132/144 = 11/12

Q:

(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 ?

A:

Total no. Of bulbs = 20

Number of defective bulbs = 4

Number of good bulbs = 16

(i) Probability of getting a defective bulb = 4/20 = 1/5

(ii) If one good bulb is kept aside,

Total no. Of bulbs = 19

Number of good bulb (not defective) = 15

Probability of getting not defective bulb = 15/19

Q:

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.

A:

Total number of discs = 90

(i) A 2-digit number discs = 81

Probability of getting two-digit number = 81/90 = 9/10

(ii) A perfect square number disc = 9

Probability of getting a perfect square number disc

= 9/90 = 1/9

(iii) A number divisible by 5 = 18

Probability of getting numbered divisible by 5

= 18/90 = 1/5

Q:

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?

A:

Total number of faces = 6

(i) A type of faces = 2

Probability of getting A type of face = 2/6 = ½

(ii) D type of face = 1

Probability of getting D type of faces = 1

Q:

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.

A:

A driver attempts to start a car. The car starts or does not Start. Not equally outcome
A player attempts to shoot a basketball. She / he shoots or misses the shot. Not equally outcome
A baby is born. It is a boy or a girl. Equally outcome

Q:

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?

A:

Area of rectangle = 3×2 = 6 m²

Area of circle = π (½)² = π/4 m²

Probability that the pie drops in the circle = (π/4) = = π/24

6

Q:

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 ?

A:

Total no. Of ball pens = 144

Number of defective ball pens = 20

Total no. Of good ball pens = 144 – 20 = 124

(i) Probability that she will get a good pen = 124/144 = 31/36

(ii) Probability that she will get a defective pen

= 20/144 = 5/36

Q:

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

Do you agree with this argument? Justify your answer.

A:

(i) Total no. Of outcomes = 36

• (1, 2) and (2, 1) are events for getting a sum as 3

P (E) = 2/36 = 1/18

• (1, 3), (2, 2) and (3, 1) are the events of getting the Sum 4

P(E) = 3/36 = 1/12

• (1, 4), (2, 3), (3, 2) and (4, 1) are the events of getting the sum 5

P(E) = 4/36 = 1/9

• (1, 5), (2, 4), (3, 3), (4, 2) and (5, 1) are the events of Getting a sum 6

P(E) = 5/36

• (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1) are the event of getting a sum 7

P(E) = 6/36 = 1/6

• (3, 6), (4, 5), (5, 4) and (6, 3) are the events of getting a sum 9

P(E) = 4/36 = 1/9

• (4, 6), (5, 5) and (6, 4) are the events of getting a sum 10

P(E) = 3/36 = 1/12

• (5, 6), (6, 5) are the events of getting a sum 11

P(E) = 2/36 =1/18

(ii) No, the eleven sum is not equally likely.

Q:

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.

A:

Total no. Of outcomes = 8

Number of outcomes if Hanif wins = 2

P(E) = 2/8 = ¼

P(not E) = 1 – ¼ = ¾

Q:

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]

A:

Total no. Of possible outcomes = 36

(i) 5 will not come either up either time = 25

P(E) = 25/36

(ii) 5 will come up at least time = 11

P(E) = 11/36

Q:

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.

A:

Total no. Of outcomes = 6

(i) P (two tails) = ¼

P (two heads) = ¼

P(one head and one tail) = 2/4 =½

So, this argument is incorrect.

(ii) P (odd no.) = 3/6 = ½

P (even no.) = 3/6 = ½

So, this statement is correct.

Q:

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?

A:

Because the outcomes of a coin head or tail are equally likely. So, this is the fair way to decide which team get the ball at the beginning.

Q:

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

(B) –1.5 (C) 15% (D) 0.7

A:

- 1.5 because probability of an event always lies between 0 and 1.

Q:

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

A:

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

0.05 + P (not E) = 1

P (Not E) = 1 – 0.05 = 0.95

Q:

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?

A:

0, impossible because there is no candy of orange flavor in a bag
1, (sure) because there are only lemon flavor candies in bag.

Q:

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?

A:

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

P (E) + 0.992 = 1

P (E) = 1 - 0.992 = 0.008

Q:

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?

A:

(i) Total no. balls = 8

Red balls = 3

Probability of red balls = 3/8

(ii) Not red balls = 5

Probability of not getting red balls = 5/8

Q:

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?

A:

Total no. of marbles = 17

(i) Total no. of red marbles = 5

Probability of getting red marbles = 5/17

(ii) Total no. of white marbles = 8

Probability of getting white marbles = 8/17

(iii) No. of not green marbles

= total no. of marbles – no. of green marbles

= 17 – 4= 13

Probability of getting not green marbles = 13/17

Q:

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?

A:

Since the number of days on which they both visit is 5 (Tuesday, Wednesday, Thurday, Friday, Saturday) they both can visit in 5 ways.

Therefore, total no of possible outcomes 5×5 = 25

(i) When the both visits the same day

Total no of favourable outcomes = 5 (Tuesday, Tuesday)

(Wednesday, Wednesday) (Thursday,Thursday) (Friday, Friday) (Saturday, Saturday)

So,P(both visiting the same day) = 5/25 = 1/5

(ii) When they visits on consecutive days 8 (Tuesday, Wednesday)

(Wednesday, Thursday) (Thursday, Friday) (Friday, Saturday) (Saturday, Friday) (Friday, Thursday)

(Thursday, Wednesday) and (Wednesday, Tuesday)

P (both visitng the consecutive days) = 8/25

(iii) when they visits the different days

P(E) = 1 – P (visits the Same day)

= 1 – 1/5 =4/5

Q:

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?

A:

+	1	2	3	4	5	6
1	2	3	3	4	4	7
2	3	4	4	5	5	8
3	3	4	4	5	5	8
4	4	5	5	6	6	9
5	4	5	5	6	6	9
6	7	8	8	9	9	12

Total no of outcomes = 6×6 = 36

(i) Even

Total no of favourable outcomes = 18

P(Even) = 18/36 = ½

(ii) Sum is 6

Total no of favourable outcomes = 4

P (sum is 6) = 4/36 =1/9

(iii) Sum is at least 6

Total no of favourable outcomes = 15

P(sum is at least 6) = 15/36 = 5/12

Q:

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.

A:

Let the number of blue balls = x

Number of red balls = 5

Therefore, total no. Of balls = 5 + x

P(E) of drawing a blue ball = [x / (5 + x)] ...........(I)

P(E) of drawing a red ball = [5 / (5 + x)]

Acc. to question,

P(E)_B = 2 P(E)_R

[x / (5 + x)] = 2 [5 / (5 + x)]

x = 10

Q:

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.

A:

Total no of balls = 12

Let the no. of black balls = x

P(E) of getting a black ball = [ x / 12]

(ii) When 6 more black balls added to bag

Total no of balls = 12 + 6 = 18

No of black balls = x + 6

P (black ball) = [(x + 6) / 18]

Acc. to question,

Before After

2 P(E)_B = P(E)_B

2 [x / 12] = [(x + 6) / 18]

x/6 = x + 6/18

x + 6 = 3x

2x = 6 => x =3

Q:

A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is 2/3.

Find the number of blue balls in the jar.

A:

Total no of marbles in the jar (green + blue) = 24

Let the no of green marbles be = x

Therefore, no of blue marbles left = 24 – x

P(E) of marble to be green = 2/3 {Given} …....(I)

Acc. to question

P(E) of green marble = x/24 …..........(ii)

Equating equation (I) and (ii)

2/3 = x/24

x = 16

No of green marbles = 16

Hence, no of blue marbles = 24- 16 = 8

Probability Question Answers: NCERT Class 10 Mathematics

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