# Maharashtra Board Class 10 Math Part 1 Solution Chapter 5 Problem Set 5 Probability

## Maharashtra Board Class 10 Math Part – 1 Solution Chapter 5 Problem Set  5 – Probability

Balbharati Maharashtra Board Class 10 Math Part – 1 Solution Chapter 5: Probability. Marathi or English Medium Students of Class 10 get here Probability full Exercise Solution.

 Std Maharashtra Class 10 Subject Math Part 1 Solution Chapter Probability Problem Set 5

### Problem set 5

(1) Choose the correct alternative answer for each of the following questions.

(1) Which number cannot represent a probability?

(A) 2/3

(B) 1.5

(C) 15%

(D) 0.7

Solution:

Which number cannot represent probability (B) 1.5 probability cannot be more than 1.

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

(A) 1/6

(B) 1/3

(C) 1/2

(D) 0

Solution:

A die is rolled. What is the probability that the numbers appearing on upper face is less than 3?

(B) 1/3

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

(A) 1/5

(B) 6/25

(C) 1/4

(D) 13/50

Solution:

What is the probability of an event that the number chosen from 1 to 100 is a prime number.

(c) 1/4, 25 prime number between 1 to 100.

(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) 1/5

(B) 3/5

(C) 4/5

(D) 1/3

Solution:

(A) 1/5, 8 multiple of 5 between 1 to 48.

(5) If n(A) = 2, P(A) = 1/5, then n (S) = ?

(A) 10

(B) 5/2

(C) 2/5

(D) 1/3

Solution:

If n (A) = 2, P (A) = 1/5, then n (s) = ?

(A) 10.

(2) Basketball players John, Vasim, Akash were practising the ball drop in the basket. The probabilities of success for John, Vasim and Akash are 4/5, 0.83 and 58% respectively. Who had the greatest probability of success?

Solution:

Probability of john = 4/5 = 4×20/5×30 = 80/100

Probability of vasim = 0.83 = 83/100

Probability of Akash = 58% = 58/100

Clearly Vasim has the most chances of success since, 83/100 > 80/100 > 58/100

(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

Solution:

There are 6 defenders, 4 offenders, 1 goalee.

∴ If s is the sample space n (s) = 6 + 4 + 1

= 11

(1) Event A: The goalee will be selected:

n(A) = 1

∴ P (A) = n (A)/n (s) = 1/11

(2) Event B: A defender will be selected.

n (B) = 6

∴ P(B) = n (B)/n (S) = 6/11

(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?

Solution:

Given, total cards = 26.

∴ If s is sample space, n (s) = 26

Event A: Total no. of vowels = {a, e, i, o, u}

∴ n (A) = 5

∴ P (A) = n (A)/n (s) = 5/26

(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

Solution:

Total no. of balls are 2 red, 3 blue, 4 green

∴ If s is the sample space.

then,  n (s) = 2 + 3 + 4 =9

(1) Event A: a red balloon

n (A) = 2

∴ P (A) = n (A)/n(s) = 2/9

(2) Event B: a blue balloon.

n (B) = 3

P (B) = n(B)/n (s) = 3/9 = 1/3

(3) Event C: a green balloon.

∴ n (c) = 4 (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?

Solution:

Total pens are 5 red, 8 blue, 3 green

∴ If s is the sample space then n (s) = 5 + 8 + 3 = 16

Event A: The pen is blue

n (A) = 8

∴ P (A) = n (A)/n (s) = 8/16 = 1/2

(7) Six face 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

Solution:

Given, are the faces of a die.

∴ s = {A, B, C, D, E, A}

n (s) = 6

(1) event A: A appears on upper face.

A = {A, A} n (A) = 2

∴ P (A) = n (A)/n (s) = 2/6 = 1/3

(2) Event B: D appears on upper face

B = {D} n {B} = 1

∴ P (B) = n (B)/n (s) = 1/6

(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

Solution:

Total no. of tickets = 30

∴ If s is sample space then, n (s) = 30

(1) Event A: odd numbers

n (A) = total no. of odd numbers between 1 to 30

= 30/2 = 15

∴ P (A) = n (A)/n (s) = 15/30 = 1/2

(2) Event B: a complete square.

B = {1, 4, 9, 10, 25}

∴ n (B) = 5

∴ p (B) = n(B)/n (s) = 5/30 = 1/6

(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.

Solution:

Length = 77m, diameter = 14m.

∴ Area of garden = 77×50 m2

∴ If s is the sample space than, n (s) = area of garden = 77×50 m2

Event A: towel fell on the circular lake.

∴ n (A) = area of circular lake = (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.

Solution:

S = {1, 2, 3, 4, 5, 6, 7, 8}

∴ n (s) = 8

(1) event A: lands on 8

A = {8}, n (A) = 1

∴ P (A) = n (A)/n (s) = 1/8

(2) Event B: lands on an odd number.

B = {1, 3, 5, 7}, n (B) = 4

∴ P (B) = n (B)/n (s) = 4/8 = 1/2

(3) Event C: lands on a number greater than 2

∴ C = {3, 4, 5, 6, 7, 8}

n (c) = 6

∴ p (c) = n (c)/n (s) = 6/8 = 3/4

(4) Event D: Lands on a number less than 9.

D = {1, 2, 3, 4, 5, 6, 7, 8}

∴ n (D) = 8

∴ P (D) = n (D)/n (S) = 8/8 = 1

(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.

Solution:

S = {0, 1, 2, 3, 4, 5}

n (s) = 6

(1) Event A: a natural number.

A = {1, 2, 3, 4, 5}

n (A) = 5

∴ P (A) = n(A)/n (s) = 5/6

(2) Event B: a number less than 1

B = {0}, n (B) = 1

∴ P (B) = n (B)/n (s) = 1/6

(3) event C: A whole numbers.

∴ C = {0, 1, 2, 3, 4, 5}

∴ n (c) = 6

∴ P (1) = n (c)/n (s) = 6/6 = 1

(4) Event D: a number greater than 5

D = {not possible}

n (D) = 0

∴ P (D) = n (D)/n (s) = 0/6 = 0

(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.

Solution:

Total no. of bells 3 red, 3 white, 3 green

∴ If s is the sample space then,

n (s) = 3 + 3 + 3 = 9

(1) Event A: a red ball

(2) n (A) = 3

∴ P (A) = n(A)/n(S) = 3/9 = 1/3

(2) Event B: n not red

n (B) = 9 – 3 = 6

∴ p (B) = n(B)/n(S) = 6/9 = 2/3

(3) Event C: Either red or white

n (c) = all red and white balls

= 3+3 = 6

∴ p (C) = n (c)/n (s) = 6/9 = 2/3

(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’.

Solution:

Each card bears was a letter from the word mathematics.

∴ s = {m, a, t, h, e, m, a, t, i, c, s}

n (s) = 11

Event A: the letter is m.

A = {m, m}

n (A) = 2

∴ P (A) = n (A)/n(s) = 2/11

(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.

Solution:

Total no. of students = 200

Students who like kabadi = 135

Students who like kabadi = 200 – 135  = 65

∴ If s is sample space than.

n (s) = 200

Event A: do not like kabadi

n (A) = 65

∴ p (A) = n (A)/n(S) = 65/200 = 13/40

(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.

Solution:

Two digit number formed by 0, 1, 2, 3, 4.

∴ S = {10, 11, 12, 13, 14, 20, 21, 22, 23, 24, 30, 31, 32, 33, 34, 40, 41, 42, 43, 44}.

n (s) = 20

(1) Event A: a prime numbers

A = {11, 13, 23, 31, 41, 43}

n (A) = 6

∴ P (A) = n(A)/n (S) = 6/20 = 3/10

(2) Event B: multiple of 4.

B = {12, 20, 24, 32, 40, 44}

n (B) = 6

∴ P (B) = n (B)/n (S) = 6/20 = 3/10

(3) Event C: multiplied of 11

C = {11, 22, 33, 44}

n (C) = 4

∴ P (C) = n (C)/n (S) = 4/20 = 1/5

(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.

Solution:

Faces of a die is 0, 1, 2, 3, 4, 5.

The die is rolled twice,

∴ S = {(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (1, 0), (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 0), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (3, 0), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (4, 0), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (5, 0), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5)}

n (s) = 36

Event A: Product of the two upper face is 0.

A = {(0, 0), (0, 2), (0, 1), (0, 3), (0, 4), (0, 5), (1, 0), (2, 0), (3, 0), (4, 0), (5, 0)}

P n (A) = 11

p (A) = n (A)/n (s) = 11/36

(17) Do the following activity –

Activity I: Total number of students in your class, n(S) =

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

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

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

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

Solution:

Activity I: n (s) = 100

n (A) = 50

A (P) = n(A)/n(S) = 50/100 = 1/2

n (B) = students not wearing specs.

= 100 – 50 = 50

∴ p (B) = 50/100 = 1/2

Activity II: Here is your solution of Maharashtra Board Class 10 Math Part – 1 Chapter 5 Probability Problem Set 5 Solution

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Updated: January 12, 2022 — 4:28 pm