ClassesClass 9MathsNCERTIntroduction to ProbabilityEnd of Chapter ExercisesQ 4
QUESTION 4 Easy

Write the sample space and calculate the probability based on the given information.
(i) Two coins are tossed at the same time. What is the probability of getting at least one head?
(ii) Two identical cards numbered 1 to 10 are placed in a box. One card is drawn at random. What is the probability of drawing a card with an even number?
(iii) A die is rolled once. What is the probability of drawing a card with an even number?
(iv) A bag contains 3 red balls, 2 blue balls, and 1 green ball. One ball is picked at random. What is the probability that it is not red?
(v) Three coins are tossed simultaneously. What is the probability of getting exactly two heads?

SOLUTION

1
(i) Two coins are tossed at the same time
Sample Space, \(S = \{HH, HT, TH, TT\}\)
Event: getting at least one head
\(E = \{HH, HT, TH\}\)
Number of favourable outcomes \(= 3\)
Total number of outcomes \(= 4\)
\(P(at\,least\,one\,head) = \frac{3}{4}\)
2
(ii) One card numbered 1 to 10 is drawn
Sample Space, \(S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\)
Event: Drawing an even number
\(E = \{2, 4, 6, 8, 10\}\)
Number of favourable outcomes \(= 5\)
Total number of outcomes \(= 10\)
\(P(even\,number) = \frac{5}{10} = \frac{1}{2}\)
3
(iii) A die is rolled once
Sample Space, \(S = \{1, 2, 3, 4, 5, 6\}\)
Event: Getting an even number
\(E = \{2, 4, 6\}\)
Number of favourable outcomes \(= 3\)
Total number of outcomes \(= 6\)
\(P(even\,number) = \frac{3}{6} = \frac{1}{2}\)
4
(iv) A bag contain s 3 red balls, 2 blue balls, and 1 green ball. One ball is picked at random
Sample Space, \(S = \{R, R, R, B, B, G\}\)
Total balls \( = 6\)
Event: Ball is not red
Non-red balls are:
\(E = \{B, B, G\}\)
Number of favourable outcomes \(= 3\)
\(P(not\,red) = \frac{3}{6} = \frac{1}{2}\)
5
Three coins are tossed simultaneously
Sample Space, \(S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}\)
Event: Getting exactly two heads
\(E = \{HHT, HTH, THH\}\)
Number of favourable outcomes \(= 3\)
Total number of outcomes \(= 8\)
\(P(exactly\,two\,heads) = \frac{3}{8}\)
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Final Answer : (i) P(at least one head) = \(\frac{3}{4}\)
(ii)P(even number) = \(\frac{1}{2}\)
(iii)P(even number) = \(\frac{1}{2}\)
(iv)P(not red) = \(\frac{1}{2}\)
P(exactly two heads) = \(\frac{3}{8}\)

Concept Note

Probability of an event is calculated using:
\(P(E) = \frac{Number\,of\,favourable\,outcomes}{Total\,number\,of\,outcomes}\), where
Sample Space (S) = all possible outcomes
Favourable outcomes(E) = outcomes required for the event

Probability always lies between 0 and 1.