HBSE 9th Class Maths Important Questions Chapter 15 Probability

Haryana State Board HBSE 9th Class Maths Important Questions Chapter 15 Probability Important Questions and Answers.

Haryana Board 9th Class Maths Important Questions Chapter 15 Probability

Very Short Answer Type Questions

Question 1.
Two coins are tossed simultaneously. What is the probability of getting at least one head?
Solution:
When two coins are tossed simultaneously all possible outcomes are HH, HT, TH, TT.
∴ Total number of possible outcomes = 4
Let E be the event of getting at least one head, then the favourable outcomes are HH, HT, TH.
Number of favourable outcomes = 3
∴ P(E) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)
= \(\frac{3}{4}\)
Hence,
∴ P(E) = \(\frac{3}{4}\)

Question 2.
Three unbaised coins are tossed simultaneously. Find the probability of getting exactly (i) 2 tails, (ii) atmost 2 tails.
Solution:
When 3 coins are tossed simultaneously all possible outcomes are HHH, HHT, THH, HTH, THT, HTT, TTH, TTT
∴ Total number of possible outcomes = 8
(i) Let E₁ be the event of getting 2 tails, then favourable outcomes are
THT, HTT, TTH
∴ Number of favourable outcomes = 3
∴ P(E₁) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)
= \(\frac{3}{8}\)

(ii) Let E₂ be the event of getting at most 2 tails, then E₂ = event of getting 0 or 1 or 2 tails
So, the favourable outcomes are HHH, HHT, THH, HTH, THT, HTT, TTH
∴ Number of favourable outcomes = 7
∴ P(E₂) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)
= \(\frac{7}{8}\)
Hence, (i) P(E₁) = \(\frac{3}{8}\), (ii) P(E₂) = \(\frac{7}{8}\)

Question 3.
A dice is thrown once. What is probability that shows as:
(i) 3,
(ii) an even number,
(iii) greater than 4.
Solution:
When a dice thrown once, all possible outcomes are : 1, 2, 3, 4, 5, 6.
∴ total number of possible outcomes = 6
(i) Let E₁ be event of getting 3, then
∴ P(E₁) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)
= \(\frac{1}{6}\)

(ii) Let E₂ be the event of getting an even number, then favourable outcomes are: 2, 4, 6.
∴ Number of favourable outcomes = 3
∴ P(E₂) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)
= \(\frac{3}{6}\) = \(\frac{1}{2}\)

(iii) Let E₃ be the event of getting a number greater than 4, then favourable outcomes are : 5,6.
∴ Number of favourable outcomes = 2
∴ P(E₃) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)
= \(\frac{2}{6}\) = \(\frac{1}{3}\)
Hence, (i) P(E₁) = \(\frac{1}{6}\)
(ii) P(E₂) = \(\frac{1}{2}\)
(ii) P(E₃) = \(\frac{1}{3}\)

Long Answer Type Questions

Question 1.
Over the past 200 working days, the number of defective parts produced by a machine is given in the following table :

Number of defective parts	Days
0	50
1	52
2	22
3	18
4	12
5	22
6	10
7	10
8	10
9	8
10	6
11	6
12	2
13	2

Determine the probability that tomorrow’s output will have
(i) no defective part
(ii) at least one defective part
(iii) not more than 5 defective parts
(iv) more than 13 defective parts.
Solution:
Total number of days = 200
∴ Total number of possible outcomes = 200
(i) Number of days having no defective parts = 50
∴ Favourable outcomes = 50
Let E₁ be the event of number of days having no defective part. Then
∴ P(E₁) = \(\frac{\text { Favourable outcomes }}{\text { Total number of possible outcomes }}\)
= \(\frac{50}{200}\) = 0.25

(ii) Number of days having at least one defective part
= 32 + 22 + 18 + 12 + 12 + 10 + 10 + 10 + 8 + 6 + 6 + 2 + 2
= 150
∴ Favourable outcomes = 150
Let E₂ be the event of number of days having at least one defective part. Then
∴ P(E₂) = \(\frac{\text { Favourable outcomes }}{\text { Total number of possible outcomes }}\)
= \(\frac{150}{200}\) = 0.75

(iii) Number of days having not more than 5 defective parts
= 50 + 32 + 22 + 18 + 12 + 12 = 146
∴ Favourable outcomes = 146.
Let E₃ be the event of number of days having not more than 5 defective parts. Then
∴ P(E₃) = \(\frac{\text { Favourable outcomes }}{\text { Total number of possible outcomes }}\)
= \(\frac{146}{200}\) = 0.73

(iv) Number of days having more than 13 defective parts = 0
∴ favourable outcomes = 0
Let E₄ be the event of number of days having more than 13 defective parts. Then
∴ P(E₄) = \(\frac{\text { Favourable outcomes }}{\text { Total number of possible outcomes }}\)
= \(\frac{0}{200}\) = 0.
Hence, (i) p(E₁) = 0·25 (ii) p(E₂) = 0.25 (iii) p(E₃) = 0.73 (iv) p(E₄) = 0.

Question 2.
Find the probability that a number selected at random from the numbers 1, 2, 3, …, 39, 40 is :
(i) a prime number less than 30.
(ii) a number which is a perfect square.
(iii) multiple of 2 or 5.
Solution:
Total number of possible outcomes = 40
(i) Out of the given numbers, prime numbers less than 30 are
2, 3, 5, 7, 11, 13, 17, 19, 23, 29
∴ Number of favourable outcomes = 10
Let E₁ be the event of getting a prime number. Then,
∴ P(E₁) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)
= \(\frac{10}{40}\) = \(\frac{1}{4}\)

(ii) Out of the given numbers, perfect square numbers are 1, 4, 9, 16, 25, 36,
∴ Number of favourable outcomes = 6
Let E₂ be the event of getting a perfect square number. Then,
∴ P(E₂) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)
= \(\frac{6}{40}\) = \(\frac{3}{20}\)

(iii) Out of the given numbers, multiple of 2 are :
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40
Multiple of 5 are: 5, 10, 15, 20, 25, 30, 35, 40
Multiple of 2 and 5 both are: 10, 20, 30, 40.
∴ Number of favourable outcomes = 20 + 8 – 4 = 24
Let E₃ be the event of getting a multiple of 2 or 5, then
∴ P(E₃) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)
= \(\frac{24}{40}\) = \(\frac{3}{5}\)
Hence, (i) P(E₁) = \(\frac{1}{4}\) (ii) P(E₂) = \(\frac{3}{20}\) (ii) P(E₃) = \(\frac{3}{5}\)

Question 3.
A bag contains 4 white balls and some blue balls. If the probability of drawing a blue ball from the bag is thrice that a white ball. Find the number of blue balls in the bag.
Solution:
Let the number of blue balls be x, then
total number of balls = 4 + x
∴ Total number of possible outcomes = 4 + x
Number of white balls = 4
∴ Number of fovourable outcomes = 4
Let E₁ be event of getting a white ball,
∴ P(E₁) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)
= \(\frac{4}{4+x}\)
Number of blue balls = x
∴ Number of favourable outcomes = x
Let E₂ be the event of getting a blue ball, then
∴ P(E₂) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)
= \(\frac{x}{4+x}\)
According to the question
P(E₂) = 3 × P(E₁)
⇒ \(\frac{x}{4+x}=3 \times \frac{4}{4+x}\)
⇒ \(\frac{x}{4+x}=\frac{12}{4+x}\)
⇒ x = \(\frac{12(4+x)}{(4+x)}\)
⇒ x = 12
Hence, number of white balls = 12.

Multiple Choice Questions

Choose the correct answer in each of the following:

Question 1.
Which one of the following cannot be probability of an event?
(a) \(\frac{1}{3}\)
(b) 0.5
(c) \(\frac{-2}{3}\)
(d) \(\frac{3}{4}\)
Answer:
(c) \(\frac{-2}{3}\)

Question 2.
Which one of the following cannot be probability of an event?
(a) \(\frac{4}{3}\)
(b) 0
(d) 0.75
(d) \(\frac{2}{5}\)
Answer:
(a) \(\frac{4}{3}\)

Question 3.
The sum of all the elementary events of an experiment is :
(a) 0
(b) 13
(c) -1
(d) \(\frac{1}{2}\)
Answer:
(b) 13

Question 4.
It is given that probability of winning a game is 0.3, then probability of losing the game is :
(a) 0.03
(b) 0.3
(c) -0.3
(d) 0.7
Answer:
(d) 0.7

Question 5.
Sum of probability of happening and not happening of an event is :
(a) 0
(b) -1
(c) 1
(d) \(\frac{1}{2}\)
Answer:
(c) 1

Question 6.
A coin is tossed 40 times and the head appears 15 times, then probability of getting a tail is :
(a) \(\frac{3}{8}\)
(b) \(\frac{1}{2}\)
(c) \(\frac{5}{8}\)
(d) \(\frac{2}{3}\)
Answer:
(c) \(\frac{5}{8}\)

Question 7.
Two coins are tossed simultaneously, Probability of getting atmost one tail is :
(a) \(\frac{1}{4}\)
(b) \(\frac{1}{2}\)
(c) \(\frac{3}{4}\)
(d) 1
Answer:
(c) \(\frac{3}{4}\)

Question 8.
In 60 throws of a dice, the outcomes were noted as given below :

Outcomes	1	2	3	4	5	6
Number of times	12	7	16	9	6	10

A dice is thrown at random. Probability of getting a not even number is :
(a) \(\frac{17}{30}\)
(b) \(\frac{13}{30}\)
(c) \(\frac{1}{5}\)
(d) \(\frac{4}{5}\)
Answer:
(a) \(\frac{17}{30}\)

Question 9.
In a family of two children, probability of at least one girl is :
(a) \(\frac{1}{4}\)
(b) \(\frac{3}{4}\)
(c) \(\frac{1}{2}\)
(d) none of these
Answer:
(b) \(\frac{3}{4}\)

Question 10.
A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. Probability that this bulb defective is :
(a) \(\frac{1}{5}\)
(b) \(\frac{1}{4}\)
(c) \(\frac{4}{5}\)
(d) \(\frac{3}{4}\)
Answer:
(a) \(\frac{1}{5}\)