Samacheer Kalvi 9th Maths Guide Chapter 9 Probability Ex 9.2

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 9 Probability Ex 9.2

Question 1.
A company manufactures 10000 Laptops in 6 months. In that 25 of them are found to be defective. When you choose one Laptop from the manufactured, what is the probability that selected Laptop is a good one?
Solution:
Total number of laptops = 10000
∴ n(S) = 10000
Number of good laptops = 10000 – 25 = 9975
Let E be the event of getting good laptops
n(E) = 9975
P(E) = \(\frac{n(E)}{n(S)}\)
= \(\frac{9975}{10000}\)
= \(\frac{399}{400}\) (or)
= 0.9975

Question 2.
In a survey of 400 youngsters aged 16-20 years, it was found that 191 have their voter ID card. If a youngster is selected at random, find the probability that the youngster does not have their voter ID card.
Solution:
Here n(S) = 400
Number of persons having voter ID = 191
Number of persons does not have their voter ID
= 400 – 191
= 209
n(E) = 209
P(E) = \(\frac{n(E)}{n(S)}\)
= \(\frac{209}{400}\)
∴ The required probability is \(\frac{209}{400}\)

Question 3.
The probability of guessing the correct answer to a certain question is \(\frac{x}{3}\). If the probability of not guessing the correct answer is \(\frac{x}{5}\), then find the value of x.
Solution:
Probability of guessing the correct answer = \(\frac{x}{3}\)
P(E) = \(\frac{x}{3}\)
Probability of not guessing the correct answer = \(\frac{x}{5}\)
P(E)’ = \(\frac{x}{5}\)
But P(E) + P(E)’ = 1
\(\frac{x}{3}\) + \(\frac{x}{5}\) = 1
\(\frac{5x+3x}{15}\) = 1 ⇒ \(\frac{8x}{15}\) = 1
8x = 15
x = \(\frac{15}{8}\)
∴ The value of x = \(\frac{15}{8}\)

Question 4.
If a probability of a player winning a particular tennis match is 0.72. What is the probability of the player loosing the match?
Solution:
Probability of a player winning a tennis match = 0.72
P(E) = 0.72
Probability of a player loosing the match be P(E)’
But P(E) + P(E)’ = 1
0.72 + P(E)’ = 1
P(E)’ = 1 – 0.72
= 0.28
Probability of the player loosing the match = 0.28

Question 5.
1500 families were surveyed and following data was recorded about their maids at homes

A family selected at random. Find the probability that the family selected has
(i) Both types of maids
(ii) Part time maids
(iii) No maids
Solution:
Total number of families surveyed = 1500
n(S) = 1500
Number of families used maids = 860 + 370 + 250
= 1480
Number of families not using any maids = 1500 – 1480
= 20

(i) Let E₁ be the event of getting families use both types of maids
n(E₁) = 250
p(E₁) = \(\frac{n(E_{1})}{n(S)}\)
= \(\frac{250}{1500}\)
= \(\frac{25}{150}\)
= \(\frac{1}{6}\)
Probability of getting both types of maids = \(\frac{1}{6}\)

(ii) Let E₂ be the event of getting families use part time maids
n(E₂) = 860
p(E₂) = \(\frac{n(E_{2})}{n(S)}\)
= \(\frac{860}{1500}\)
= \(\frac{43}{75}\)
Probability of getting part time maids = \(\frac{43}{75}\)

(iii) Let E₃ be the event of getting family use no maids
n(E₃) = 20
p(E₃) = \(\frac{n(E_{3})}{n(S)}\)
= \(\frac{20}{1500}\)
= \(\frac{2}{150}\)
= \(\frac{1}{75}\)
Probability of getting no maids = \(\frac{1}{75}\)