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?
Step-by-step solution
Idea: There are 5 cards, so 5 equally likely outcomes. E is on 2 cards, so count it twice.
(i) What is the probability that it is a P, E or C?
- Cards showing P, E or C: P (1), E (2), C (1): 4 cards. Only A is left out.½ mark
- P = 45 = 0.8.½ mark
(ii) What is the probability that it is not an E?
- Cards that are not E: P, A, C: 3 cards.½ mark
- P(not E) = 35 = 0.6. (Check: P(E) = 25 and 1 − 25 = 35.)½ mark
Check: (i) P(A) = 15 and 45 + 15 = 1 ✓. (ii) P(E) = 25 and 35 + 25 = 1 ✓.
Answer to write in the exam
(i)
n(S) = 5 (P, E, A, C, E)
Favourable cards: P, E, C, E; number = 4
∴ P(P, E or C) = 45
(ii)
Cards not E: P, A, C; number = 3
∴ P(not E) = 35
Common mistakes that cost marks
- Counting E only once and using 4 cards in all. PEACE has 5 letters, and both Es are separate cards.
- In (i), getting 35 by counting P, E, C as three cards. E is on two cards, so there are 4 favourable cards.
- In (ii), getting 45 by removing only one E.
How this can come in the exam
The letters of the word ‘APPLE’ are written on 5 cards and one is drawn at random. The probability that it is P is
- 15
- 25
- 14
- 12
Show answer
(B) 25
P appears on 2 of the 5 cards: 25.
The letters of the word ‘COCONUT’ are put on cards and one card is drawn at random. Find the probability that the letter is (a) C or O, (b) not a vowel.
Show answer
7 cards: C, O, C, O, N, U, T. (a) C, C, O, O: P = 47. (b) Vowels O, O, U; not a vowel: C, C, N, T: P = 47.Try one yourself
The letters of the word ‘SEED’ are placed on cards. One card is drawn at random. Find the probability that it is not an E.
Show answer
4 cards, 2 are E. P(not E) = 24 = 12.
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