QUESTION 8
Easy
The letters of the word 'PEACE' are placed on cards. Leela draws a card without looking.
(i) What is the probability that it is a P, E, or C?
(ii) What is the probability that it is not an E?
SOLUTION
1
Write the sample space
The letters in the word PEACE are:
\(S = \{P, E, A, C, E\}\)
Total number of cards, \(n(S) = 5\)
The letter \(E\) appears twice, so both E's are counted separately.
\(S = \{P, E, A, C, E\}\)
Total number of cards, \(n(S) = 5\)
The letter \(E\) appears twice, so both E's are counted separately.
2
(i) Probability of drawing P, E, or C
Since E appears twice, the favourable outcomes are:
\(E = \{P, E, C, E\}\)
Number of favourable outcomes, \(n(E) = 4\)
So, \(P(E) = \frac{4}{5}\)
The probability that it is a P, E, or C is \(\frac{4}{5}\).
\(E = \{P, E, C, E\}\)
Number of favourable outcomes, \(n(E) = 4\)
So, \(P(E) = \frac{4}{5}\)
The probability that it is a P, E, or C is \(\frac{4}{5}\).
3
(ii) Probability of drawing a letter that is not E
Letters other than E are P, A, and C
\(E = \{P, A, C\}\)
Number of favourable outcomes, \(n(E) = 3\)
So, \(P(E) = \frac{3}{5}\) The probability that it is not an E is \(\frac{3}{5}\).
\(E = \{P, A, C\}\)
Number of favourable outcomes, \(n(E) = 3\)
So, \(P(E) = \frac{3}{5}\) The probability that it is not an E is \(\frac{3}{5}\).
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Final Answer :
(i) P(P, E or C) = \(\frac{4}{5}\)
(ii) P(not E) =\(\frac{3}{5}\)
Concept Note
The sample space contains all possible outcomes.
Repeated letters are counted separately.