Mathematics Part i Solutions Solutions for Class 10 Maths Chapter 5 - Probability

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Page No 116:

Question 1:

How many possibilities are there in each of the following?
(1) Vanita knows the following sites in Maharashtra. She is planning to visit one of them in her summer vacation.
Ajintha, Mahabaleshwar, Lonar Sarovar, Tadoba wild life sanctuary, Amboli, Raigad, Matheran, Anandavan.
(2) Any day of a week is to be selected randomly.
(3) Select one card from the pack of 52 cards.
(4) One number from 10 to 20 is written on each card. Select one card randomly.

Answer:

(1) Total number of possible palces in Maharashtra which can Vanita visits = 8
(2) Total number of days in a week = Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday = 7
(3) Total number of cards = 52
(4) Total number of cards = 11

Page No 117:

Question 1:

For each of the following experiments write sample space ‘S’ and number of sample points n(S).
(1) One coin and one die are thrown simultaneously.
(2) Two digit numbers are formed using digits 2, 3 and 5 without repeating a digits.

Answer:

(1) Sample Space (S) = {1H, 1T, 2H, 2T, 3H, 3T, 4H, 4T, 5H, 5T, 6H, 6T}
∴ n(S) = 12
Hence, the required number of sample spaces is 12.
(2) Sample Space (S) = {23, 25, 32, 35, 52, 53}
∴ n(S) = 6
Hence, the required number of sample spaces is 6.

Page No 118:

Question 2:

The arrow is rotated and it stops randomly on the disc. Find out on which colour it may stop.

Answer:

Sample Space (S) = {Red, Purple, Orange, Yellow, Blue, Green}
∴ n(S) = 6
Hence, the number of colours on which the arrow may stop is 6.

Page No 118:

Question 3:

In the month of March 2019, Find the days on which the date is a multiple of 5. (see the given page of the calender)

Answer:

The dates multiple of 5 are 5, 10, 15, 20, 25 and 30
From the calender, we can observe on these dates, the days fall are Tuesday, Sunday, Friday, Wednesday, Monday, Saturday
Sample Space = {Tuesday, Sunday, Friday, Wednesday, Monday, Saturday}
∴ n(S) = 6
Hence, the total number days on which the date is a multiple of 5 is 6.

Page No 118:

Question 4:

Form a ‘Road safety commitee’ of two, from 2 boys (B₁, B₂) and 2 girls(G₁, G₂). Complete the following activity to write the sample space.
(a) Committee of 2 boys = $x$

(b) Committee of 2 girls = $x$

(c) Committee of one boy and one girl = $B_{1} G_{1}$ $x$ $x$ $x$

∴ Sample space = {..., ..., ..., ..., ..., ...}

Answer:

(a) Committee of 2 boys = B₁, B₂

(b) Committee of 2 girls = G₁, G₂

(c) Committee of one boy and one girl = (B₁, G₁), (B₁, G₂), (B₂, G₁), (B₂, G₂)

∴ Sample space = {(B₁, B₂), (G₁, G₂), (B₁, G₁), (B₁, G₂), (B₂, G₁), (B₂, G₂)}
or n(S) = 6

Page No 121:

Question 1:

Write sample space ‘S’ and number of sample point n(S) for each of the following experiments. Also write events A, B, C in the set form and write n(A), n(B), n(C).
(1) One die is rolled,
Event A : Even number on the upper face.
Event B : Odd number on the upper face.
Event C : Prime number on the upper face.

(2) Two dice are rolled simultaneously,
Event A : The sum of the digits on upper faces is a multiple of 6.
Event B : The sum of the digits on the upper faces is minimum 10.
Event C : The same digit on both the upper faces.

(3) Three coins are tossed simultaneously.
Condition for event A : To get at least two heads.
Condition for event B : To get no head.
Condition for event C : To get head on the second coin.

(4) Two digit numbers are formed using digits 0, 1, 2, 3, 4, 5 without repetition of the digits.
Condition for event A : The number formed is even
Condition for event B : The number formed is divisible by 3.
Condition for event C : The number formed is greater than 50.

(5) From three men and two women, environment committee of two persons is to be formed.
Condition for event A : There must be at least one woman member.
Condition for event B : One man, one woman committee to be formed.
Condition for event C : There should not be a woman member.

(6) One coin and one die are thrown simultaneously.
Condition for event A : To get head and an odd number.
Condition for event B : To get a head or tail and an even number.
Condition for event C : Number on the upper face is greater than 7 and tail on the coin.

Answer:

(1) Sample Space = {1, 2, 3, 4, 5, 6}
∴ n(S) = 6
A = {2, 4, 6}
∴ n(A) = 3,
B = {1, 3, 5}
∴ n(B) = 3,
C = {2, 3, 5}
∴ n(C) = 3
(2) $Sample Space (S) = \{\begin{matrix} (1,1) & (1, 2) & (1, 3) & (1, 4) & (1, 5) & (1, 6) \\ (2,1) & (2, 2) & (2, 3) & (2, 4) & (2, 5) & (2, 6) \\ (3,1) & (3, 2) & (3, 3) & (3, 4) & (3, 5) & (3, 6) \\ (4,1) & (4, 2) & (4, 3) & (4, 4) & (4, 5) & (4, 6) \\ (5,1) & (5, 2) & (5, 3) & (5, 4) & (5, 5) & (5, 6) \\ (6, 1) & (6, 2) & (6, 3) & (6, 4) & (6, 5) & (6, 6) \end{matrix}\}$

∴ n(S) = 36
A = {(1, 5) (2, 4) (3, 3) (4, 2) (5, 1) (6, 6)}
∴ n(A) = 6

B = {(4, 6) (5, 5) (5, 6) (6, 4) (6, 5) (6, 6)}
∴ n(B) = 6
C = {(1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)}
∴ n(C) = 6

(3) Sample Space (S) = {HHH, HHT, HTT, HTH, THT, TTH, THH, TTT}
∴ n(S) = 8
A = {HHH, HHT, HTH, THH}
∴ n(A) = 4
B = {TTT}
∴ n(B) = 1
C = {HHH, HHT, THH}
∴ n(C) = 3

(4) Sample Space (S) = {10, 12, 13, 14, 15, 20, 21, 23, 24, 25, 30, 31, 32, 34, 35, 40, 41, 42, 43,
45, 50, 51, 52, 53, 54}
∴ n(S) = 25
A = {10, 12, 14, 20, 24, 30, 32, 34, 40, 42, 50, 52, 54}
∴ n(A) = 13
B = {12, 15, 21, 24, 30, 42, 45, 51, 54}
∴ n(B) = 9
C = {51, 52, 53, 54}
∴ n(C) = 4

(5) S = {M₁M₂, M₁M₃, M₁F₁, M₁F₂, M₂M₃, M₂F₁, M₂F₂, M₃F₁, M₃F₂, F₁F₂}
∴ n(S) = 10
A = {M₁F₁, M₁F₂, M₂F₁, M₂F₂, M₃F₁, M₃F₂, F₁F₂}
∴ n(A) = 7
B = {M₁F₁, M₁F₂, M₂F₁, M₂F₂, M₃F₁, M₃F₂}
n(B) = 6
C = {M₁M₂, M₁M₃, M₂M₃, }
∴ n(C) = 3

(6) S = {H1, H2, H3, H4, H5, H6 T1, T2, T3, T4, T5, T6}
∴ n(S) = 12
A = {H1, H3, H5}
∴ n(A) = 3
B = {H2, H4, H6, T2, T4, T6}
∴ n(B) = 6
C = {}
∴ n(C) = 0

Page No 125:

Question 1:

If two coins are tossed, find the probability of the following events.
(1) Getting at least one head.
(2) Getting no head.

Answer:

Sample Space (S) = {HH, HT, TT, TH}
∴ n(S) = 4
(1) Event A: Getting at least one head
A = {HH, HT, TH}
∴ n(A) = 3
$∴ P (A) = \frac{n (A)}{n (S)} = \frac{3}{4}$
Hence, the probability of getting at least one head is $\frac{3}{4}$ .
(2) Event B: Getting no head
B = {TT}
∴ n(B) = 1
$∴ P (B) = \frac{n (B)}{n (S)} = \frac{1}{4}$
â€‹Hence, the probability of getting no head is $\frac{1}{4}$ .

Page No 125:

Question 2:

If two dice are rolled simultaneously, find the probability of the following events.
(1) The sum of the digits on the upper faces is at least 10.
(2) The sum of the digits on the upper faces is 33.
(3) The digit on the first die is greater than the digit on second die.

Answer:

$Sample Space (S) = \{\begin{matrix} (1,1) & (1, 2) & (1, 3) & (1, 4) & (1, 5) & (1, 6) \\ (2,1) & (2, 2) & (2, 3) & (2, 4) & (2, 5) & (2, 6) \\ (3,1) & (3, 2) & (3, 3) & (3, 4) & (3, 5) & (3, 6) \\ (4,1) & (4, 2) & (4, 3) & (4, 4) & (4, 5) & (4, 6) \\ (5,1) & (5, 2) & (5, 3) & (5, 4) & (5, 5) & (5, 6) \\ (6, 1) & (6, 2) & (6, 3) & (6, 4) & (6, 5) & (6, 6) \end{matrix}\}$
∴ n(S) = 36

(1) Event A: sum of the digits on the upper faces is at least 10
A = {(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)}
∴ n(A) = 6
$∴ P (A) = \frac{n (A)}{n (S)} = \frac{6}{36} = \frac{1}{6}$
Hence, the probability that the sum of the digits on the upper faces is at least 10 is $\frac{1}{6}$ .
(2) Event B: sum of the digits on the upper faces is 33
B = {}
∴ n(B) = 0
$∴ P (B) = \frac{n (B)}{n (S)} = \frac{0}{36}$
= 0
â€‹Hence, the probability that the sum of the digits on the upper faces is 33 is 0.

(3) Event C: digit on the first die is greater than the digit on second die
C = {(2,1), (3,1), (3,2), (4,1), (4,2), (4,3), (5,1), (5,2), (5,3), (5,4), (6,1), (6,2), (6,3), (6,4), (6,5)}
∴ n(C) = 15
$∴ P (C) = \frac{n (C)}{n (S)} = \frac{15}{36} = \frac{5}{12}$
â€‹Hence, the probability that digit on the first die is greater than the digit on second die is $\frac{5}{12}$ .

Page No 125:

Question 3:

There are 15 tickets in a box, each bearing one of the numbers from 1 to 15. One ticket is drawn at random from the box. Find the probability of event that the ticket drawn -
(1) shows an even number.
(2) shows a number which is a multiple of 5.

Answer:

Sample Space (S) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
∴ n(S) = 15

(1) Event A: the ticket shows an even number
A = {2, 4, 6, 8, 10, 12, 14}
∴ n(A) = 7
$∴ P (A) = \frac{n (A)}{n (S)} = \frac{7}{15}$
Hence, the probability that the ticket shows an even number is $\frac{7}{15}$ .
(2) Event B: the ticket shows a number which is a multiple of 5
B = {5, 10, 15 }
∴ n(B) = 3
$∴ P (B) = \frac{n (B)}{n (S)} = \frac{5}{15} = \frac{1}{5}$
â€‹Hence, the probability that the ticket shows a number which is a multiple of 5 is $\frac{1}{5}$ .

Page No 125:

Question 4:

A two digit number is formed with digits 2, 3, 5, 7, 9 without repetition. What is the probability that the number formed is
(1) an odd number ?
(2) a multiple of 5 ?

Answer:

Sample Space (S) = {23, 25, 27, 29, 32, 35, 37, 39, 52, 53, 57, 59, 72, 73, 75, 79, 92, 93, 95, 97}
∴ n(S) = 20

(1) Event A: the number is odd
A = {23, 25, 27, 29, 35, 37, 39, 53, 57, 59, 73, 75, 79, 93, 95, 97}
∴ n(A) = 16
$∴ P (A) = \frac{n (A)}{n (S)} = \frac{16}{20} = \frac{4}{5}$
Hence, the probability that the number is odd is $\frac{1}{6}$ .
(2) Event B: the number is a multiple of 5
B = {25, 35, 75, 95}
∴ n(B) = 4
$∴ P (B) = \frac{n (B)}{n (S)} = \frac{4}{20} = \frac{1}{5}$
â€‹Hence, the probability that the number is a multiple of 5 is $\frac{1}{5}$ .

Page No 125:

Question 5:

A card is drawn at random from a pack of well shuffled 52 playing cards. Find the probability that the card drawn is –
(1) an ace.
(2) a spade.

Answer:

(1)
Event A: the card drawn is an ace
∴ n(A) = 4
$∴ P (A) = \frac{n (A)}{n (S)} = \frac{4}{52} = \frac{1}{13}$
Hence, the probability that the card drawn is an ace is $\frac{1}{13}$ .
(2)
Event B: the card drawn is a spade
∴ n(B) = 13
$∴ P (B) = \frac{n (B)}{n (S)} = \frac{13}{52} = \frac{1}{4}$
Hence, the probability that the card drawn is a spade is $\frac{1}{4}$ .

Page No 126:

Question 1:

Choose the correct alternative answer for each of the following questions.
(1)Which number cannot represent a probability ?

(A)

\frac{2}{3}

(B) 1.5

(D) 0.7

(2) A die is rolled. What is the probability that the number appearing on upper face is less than 3 ?

(A)

\frac{1}{6}

(B)

\frac{1}{3}

(C)

\frac{1}{2}

(D) 0

(3) What is the probability of the event that a number chosen from 1 to 100 is a prime number ?

(A)

\frac{1}{5}

(B)

\frac{6}{25}

(C)

\frac{1}{4}

(D)

\frac{13}{50}

(4) There are 40 cards in a bag. Each bears a number from 1 to 40. One card is drawn at random. What is the probability that the card bears a number which is a multiple of 5 ?

(A)

\frac{1}{5}

(B)

\frac{3}{5}

(C)

\frac{4}{5}

(D)

\frac{1}{3}

(5) If n(A) = 2, P(A) =

\frac{1}{5}

, then n(S) = ?

(A) 10

(B)

\frac{5}{2}

(C)

\frac{2}{5}

(D)

\frac{1}{3}

Answer:

(1)
The probability cannot be greater than 1.
Hence, the correct option is (B).
(2)
Sample Space (S) = {1, 2, 3, 4, 5, 6}
∴ n(S) = 6
Event A: a number is less than 3
A = {1, 2}
∴ n(A) = 2
$∴ P (A) = \frac{n (A)}{n (S)} = \frac{2}{6} = \frac{1}{3}$
â€‹Hence, the correct option is (B).

(3)
Sample Space (S) = {1, 2, ......, 100}
∴ n(S) = 100
Event A: a prime number
A = {2, 3, 5,.....97}
∴ n(A) = 25
$∴ P (A) = \frac{n (A)}{n (S)} = \frac{25}{100} = \frac{1}{4}$
â€‹Hence, the correct option is (C).

(4)
Sample Space (S) = {1, 2, ......, 40}
∴ n(S) = 40
Event A: a multiple of 5
A = {5, 10, 15, 20, 25, 30, 35, 40}
∴ n(A) = 8
â€‹ $∴ P (A) = \frac{n (A)}{n (S)} = \frac{8}{40} = \frac{1}{5}$
â€‹Hence, the correct option is (A).

(5)
$P (A) = \frac{n (A)}{n (S)} \Rightarrow n (S) = \frac{n (A)}{P (A)} = \frac{2}{\frac{1}{5}}$
= 10
â€‹Hence, the correct option is (A).

Page No 126:

Question 3:

In a hockey team there are 6 defenders , 4 offenders and 1 goalee. Out of these, one player is to be selected randomly as a captain. Find the probability of the selection that –
(1) The goalee will be selected.
(2) A defender will be selected.

Answer:

Given:
Number of defenders = 6
Number of offenders = 4
Number of goalee = 1
Let S be the sample space.
∴ n(S) = Number of defenders + Number of offenders + Number of goalee
= 6 + 4 + 1
= 11
(1) Event A: The goalee will be selected
Given: n(A) = 1
$∴ P (A) = \frac{n (A)}{n (S)} = \frac{1}{11}$
Hence, the probability that the goalee will be selected is $\frac{1}{9}$ .
(2) Event B: A defender will be selected
Given: n(B) = 6
$∴ P (B) = \frac{n (B)}{n (S)} = \frac{6}{11}$
Hence, the probability that the goalee will be selected is $\frac{6}{11}$ .

Page No 126:

Question 4:

Joseph kept 26 cards in a cap, bearing one English alphabet on each card. One card is drawn at random. What is the probability that the card drawn is a vowel card ?

Answer:

Let S be the sample space.
We are given that there are 26 cards in a cap bearing one English alphabet on each card.
∴ n(S) = 26
Let A be the event of drawing is a vowel card.
There are 5 vowels out of 26 English alphabets.
∴ n(A) = 5
$∴ P (A) = \frac{n (A)}{n (S)} = \frac{5}{26}$
Hence, the probability that the card drawn is a vowel card is $\frac{5}{26}$ .

Page No 126:

Question 5:

A balloon vendor has 2 red, 3 blue and 4 green balloons. He wants to choose one of them at random to give it to Pranali. What is the probability of the event that Pranali gets,
(1) a red balloon
(2) a blue balloon
(3) a green balloon.

Answer:

Given:
Number of red baloon = 2
Number of blue baloon = 3
Number of green baloon = 4
Let S be the sample space.
∴ n(S) = Number of red baloon + Number of blue balloon + Number of green balloon
= 2 + 3 + 4
= 9
(1) Let A be the event of getting a red balloon.
Given: n(A) = 2
$∴ P (A) = \frac{n (A)}{n (S)} = \frac{2}{9}$
Hence, the probability of the event that Pranali gets a red baloon is $\frac{2}{9}$ .
(2) Let B be the event of getting a blue balloon.
Given: n(B) = 3
$∴ P (B) = \frac{n (B)}{n (S)} = \frac{3}{9} = \frac{1}{3}$
Hence, the probability of the event that Pranali gets a blue baloon is $\frac{1}{3}$ .
(3)
Let C be the event of getting a green balloon.
Given: n(C) = 4
$∴ P (C) = \frac{n (C)}{n (S)} = \frac{4}{9}$
Hence, the probability of the event that Pranali gets a green baloon is $\frac{4}{9}$ .

Page No 127:

Question 6:

A box contains 5 red, 8 blue and 3 green pens. Rutuja wants to pick a pen at random. What is the probability that the pen is blue?

Answer:

Total number of pens = 5 + 8 + 3
= 16
Number of blue pens = 8
Therefore, required probability = $\frac{Number of blue pens}{Total number of pens}$
$= \frac{8}{16} = \frac{1}{2}$
Hence, the probability that the pen is blue is $\frac{1}{2}$ .

Page No 127:

Question 7:

Six faces of a die are as shown below.

If the die is rolled once, find the probability of -
(1) ‘A’ appears on upper face.
(2) ‘D’ appears on upper face.

Answer:

Total number of faces = 6
(1)
Number of faces having A = 2
Therefore, required probability = $\frac{Number of faces having A}{Total number of faces}$
$= \frac{2}{6} = \frac{1}{3}$
Hence, the probability that ‘A’ appears on upper face is $\frac{1}{3}$ .
(2)
Number of faces having D = 1
Therefore, required probability = $\frac{Number of faces having D}{Total number of faces}$
$= \frac{1}{6}$
Hence, the probability that ‘D’ appears on upper face is $\frac{1}{6}$ .

Page No 127:

Question 8:

A box contains 30 tickets, bearing only one number from 1 to 30 on each. If one ticket is drawn at random, find the probability of an event that the ticket drawn bears
(1) an odd number
(2) a complete square number.

Answer:

Total number of tickets = 30
(1)
Number of tickets bearing odd number = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29
= 15
Therefore, required probability = $\frac{Number of tickets bearing odd number}{Total number of tickets}$
$= \frac{15}{30} = \frac{1}{2}$
Hence, the probability that the ticket drawn bears an odd number is $\frac{1}{2}$ .
(2)
Number of tickets bearing perfect square number = 1, 4, 9, 16, 25
= 5
Therefore, required probability = $\frac{Number of tickets bearing perfect number}{Total number of tickets}$
$= \frac{5}{30} = \frac{1}{6}$
Hence, the probability that the ticket drawn bears a perfect square number is $\frac{1}{6}$ .

Page No 127:

Question 9:

Length and breadth of a rectangular garden are 77 m and 50 m. There is a circular lake in the garden having diameter 14 m. Due to wind, a towel from a terrace on a nearby building fell into the garden. Then find the probability of the event that it fell in the lake.

Answer:

Area of the garden = 77 x 50
= 3850
Diameter of the circular lake = 14 m
∴ Radius of the circular lake = $\frac{14}{2}$
= 7 m
Now, area of the circular lake = $π {(radius)}^{2}$
= $\frac{22}{7} \times 7 \times 7$
= 154
Therefore, required probability = $\frac{154}{3850}$
$= \frac{1}{25}$
Hence, the probability that towel fell in the lake is $\frac{1}{25}$ .

Page No 127:

Question 10:

In a game of chance, a spinning arrow comes to rest at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8. All these are equally likely outcomes. Find the probability that it will rest at

(1) 8.
(2) an odd number.
(3) a number greater than 2.
(4) a number less than 9

Answer:

Total number of outcomes = 8
(1) Number of favourable outcomes (8) = 1
Therefore, required probability = $\frac{Number of favorable outcomes}{Total number of outcomes}$
$= \frac{1}{8}$
Hence, the probability that the spinning arrow will rest at 8 is $\frac{1}{8}$ .
(2) Number of favourable outcomes (odd number) = 1, 3, 5, 7
= 4
Therefore, required probability = $\frac{Number of favorable outcomes}{Total number of outcomes}$
$= \frac{4}{8} = \frac{1}{2}$
Hence, the probability that the spinning arrow will rest at odd number is $\frac{1}{2}$ .
(3) Number of favourable outcomes (a number greater than 2) = 3, 4, 5, 6, 7 and 8
= 6
Therefore, required probability = $\frac{Number of favorable outcomes}{Total number of outcomes}$
$= \frac{6}{8} = \frac{3}{4}$
Hence, the probability that the spinning arrow will rest at a number greater than 2 is $\frac{3}{4}$ .
(4) Number of favourable outcomes (a number less than 9) = 1, 2, 3, 4, 5, 6, 7, 8
= 8
Therefore, required probability = $\frac{Number of favorable outcomes}{Total number of outcomes}$
$= \frac{8}{8}$
= 1
Hence, the probability that the spinning arrow will rest at a number less than 9 is 1.

Page No 127:

Question 11:

There are six cards in a box, each bearing a number from 0 to 5. Find the probability of each of the following events, that a card drawn shows,
(1) a natural number.
(2) a number less than 1.
(3) a whole number.
(4) a number is greater than 5.
figure

Answer:

Total number of cards = 6
(1)
Number of cards bearing a natural number = 1, 2, 3, 4, 5
= 5
Therefore, required probability = $\frac{Number of cards bearing a natural number}{Total number of cards}$
$= \frac{5}{6}$
Hence, the probability that a card drawn shows a natural number is $\frac{5}{6}$ .
(2)
Number of cards bearing a number less than 1 = 1
Therefore, required probability = $\frac{Number of cards bearing a number less than 1}{Total number of cards}$
$= \frac{1}{6}$
Hence, the probability that a card drawn shows a number less than 1 is $\frac{1}{6}$ .
(3)
Number of cards bearing a whole number = 0, 1, 2, 3, 4, 5
= 6
Therefore, required probability = $\frac{Number of cards bearing a whole number}{Total number of cards}$
$= \frac{6}{6}$
= 1
Hence, the probability that a card drawn shows a whole number is 1.
(4)
Number of cards bearing a number greater than 5 = 0
Therefore, required probability = $\frac{Number of cards bearing a number greater than 5}{Total number of cards}$
$= \frac{0}{6}$
= 0
Hence, the probability that a card drawn shows a number greater than 5 is 0.

Page No 128:

Question 12:

A bag contains 3 red, 3 white and 3 green balls. One ball is taken out of the bag at random. What is the probability that the ball drawn is –
(1) red.
(2) not red
(3) either red or white.

Answer:

Total number of balls = 3 + 3 + 3
= 9
(1) Number of red balls = 3
Therefore, required probability = $\frac{Number of red balls}{Total number of balls}$
$= \frac{3}{9} = \frac{1}{3}$
Hence, the probability that a red ball is drawn is $\frac{1}{3}$ .
(2) Number of balls which are not red = 3 + 3
Therefore, required probability = $\frac{Number of not red balls}{Total number of balls}$
$= \frac{6}{9} = \frac{2}{3}$
Hence, the probability that a ball drawn is not red is $\frac{2}{3}$ .
(3) Number of red and white balls = 3 + 3
Therefore, required probability = $\frac{Number of red and white balls}{Total number of balls}$
$= \frac{6}{9} = \frac{2}{3}$
Hence, the probability that a ball drawn is either red or white is $\frac{2}{3}$ .

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Question 13:

Each card bears one letter from the word ‘mathematics’ The cards are placed on a table upside down. Find the probability that a card drawn bears the letter ‘m’.

Answer:

Total number of letters in word 'mathematics' = 11
= 36
Number of letters in word 'mathematics' = 2
Therefore, required probability = $\frac{Number of letters m}{Total number of letters}$
$= \frac{2}{11}$
Hence, the probability that a card drawn bears the letter ‘m’ is $\frac{2}{11}$ .

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Question 14:

Out of 200 students from a school, 135 like Kabbaddi and the remaining students do not like the game. If one student is selected at random from all the students, find the probability that the student selected dosen't like Kabbaddi.

Answer:

Total number of students = 200
Number of students who like Kabbaddi = 135
Therefore, number of students who don't like Kabbaddi = 200 − 135
= 65
Therefore, required probability = $\frac{Number of students who don' t like Kabbaddi}{Total number of students}$
$= \frac{65}{200} = \frac{13}{40}$
Hence, the probability that the student selected dosen't like Kabbaddi is $\frac{13}{40}$ .

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Question 15:

A two digit number is to be formed from the digits 0, 1, 2, 3, 4. Repetition of the digits is allowed. Find the probability that the number so formed is a -
(1) prime number
(2) multiple of 4
(3) multiple of 11.

Answer:

Two digits numbers which can be formed using the digits 0, 1, 2, 3, 4 are 10, 11, 12 ,13 ,14 ,20, 21 ,22, 23, 24, 30, 31, 32, 33, 34, 40, 41, 42, 43 and 44
Therefore, total numbers outcomes = 20
(1) Number of favourable outcomes (prime numbers) = 11, 13, 23, 31, 41 and 43
= 6
Therefore, required probability = $\frac{Number of favorable outcomes}{Total number of outcomes}$
$= \frac{6}{20} = \frac{3}{10}$ â€‹
Hence, the probability that the prime number formed is $\frac{3}{10}$ .

(2) Number of favourable outcomes (multiple of 4) = 12, 20, 24, 32, 40 and 44
= 6
Therefore, required probability = $\frac{Number of favorable outcomes}{Total number of outcomes}$
â€‹ $= \frac{6}{20} = \frac{3}{10}$ â€‹
Hence, the probability that the number so formed is a multiple of 4 is $\frac{3}{10}$ .
(3) Number of favourable outcomes (multiple) = 11, 22, 33 and 44
= 4
Therefore, required probability = $\frac{Number of favorable outcomes}{Total number of outcomes}$
â€‹ $= \frac{4}{20} = \frac{1}{5}$ â€‹
â€‹Hence, the probability that the number so formed is a multiple of 11 is $\frac{1}{5}$ .

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Question 16:

The faces of a die bear numbers 0, 1, 2, 3, 4, 5. If the die is rolled twice, then find the probability that the product of digits on the upper face is zero.

Answer:

Total number of outcomes = 6²
= 36
Number of favourable outcomes (product of digits on the upper face is zero) = (0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (1, 0), (2, 0), (3, 0), (4, 0) and (5, 0)
= 11
Therefore, required probability = $\frac{Number of favorable outcomes}{Total number of outcomes}$
$= \frac{11}{36}$
â€‹Hence, the probability that the product of digits on the upper face is zero is $\frac{3}{10}$ .

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Question 17:

Do the following activity -
Activity I : Total number of students in your class, n(S) = $x$

Number of students from your class, wearing spectacles, n(A) = $x$

Probability of a randomly selected student wearing spectacles, P(A) = $x$

Probability of a randomly selected student not wearing spectacles, P(B) = $x$

Activity II : Decide the sample space yourself and fill in the following boxes.

Answer:

Activity I :

Total number of students in your class, n(S) = 40

Number of students from your class, wearing spectacles, n(A) = 6

Probability of a randomly selected student wearing spectacles, P(A) = $\frac{6}{40}$
$= \frac{3}{20}$

Probability of a randomly selected student not wearing spectacles, P(B) = $\frac{40 - 6}{40}$
$= \frac{34}{40} = \frac{17}{20}$

Activity II :
Let us suppose the sample space be {1, 2, 3, 4, 5, 6, 7, 8, 9 10}
Event A: Getting an even number
A = {2, 4, 6, 8, 10}
∴ n(A) = 5
Now, $P (A) = \frac{n (A)}{n (S)}$

$= \frac{5}{10} = \frac{1}{2}$

View NCERT Solutions for all chapters of Class 10