(a) A pair of fair dice each numbered 1 to 6 is tossed. Find the probability of getting a sum of at least 9.
(b) If the probability that a civil servant owns a car is \(\frac{1}{6}\), find the probability that:
(i) two civil servants, A and B, selected at random each owns a car ; (ii) of two civil servants, C and D selected at random, only one owns a car ; (iii) of three civil servants, X, Y and Z, selected at random, only one owns a car.
(b) If the probability that a civil servant owns a car is \(\frac{1}{6}\), find the probability that:
(i) two civil servants, A and B, selected at random each owns a car ; (ii) of two civil servants, C and D selected at random, only one owns a car ; (iii) of three civil servants, X, Y and Z, selected at random, only one owns a car.
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Correct Answer: Option n
Explanation:
(a) Sample space for a pair of fair dice = 36 results
Total with sum at least 9 = 10
P(sum at least 9) = \(\frac{10}{36} = \frac{5}{18}\)
(b) P(civil servant owns a car) = \(\frac{1}{6}\)
P(civil servant owns no car) = \(\frac{5}{6}\)
(i) P(A and B own a car) = \(\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\)
(ii) P(C and D, only one owns a car) = P(C owns, not D) + P(not C but D owns)
= \((\frac{1}{6} \times \frac{5}{6}) + (\frac{5}{6} \times \frac{1}{6})\)
= \(\frac{5}{36} + \frac{5}{36}\)
= \(\frac{10}{36} = \frac{5}{18}\)
(iii) P(of X, Y and Z, only one owns a car) = P(X owns a car, not Y, not Z) + P(not X, Y owns a car, not Z) + P(not X, not Y, Z owns a car)
= \(3(\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6})\)
= \(3 \times \frac{25}{216}\)
= \(\frac{25}{72}\)
(a) Sample space for a pair of fair dice = 36 results
| + | 1 | 2 | 3 | 4 | 5 | 6 |
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| 6 | 7 | 8 | 9 | 10 | 11 | 12 |
Total with sum at least 9 = 10
P(sum at least 9) = \(\frac{10}{36} = \frac{5}{18}\)
(b) P(civil servant owns a car) = \(\frac{1}{6}\)
P(civil servant owns no car) = \(\frac{5}{6}\)
(i) P(A and B own a car) = \(\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\)
(ii) P(C and D, only one owns a car) = P(C owns, not D) + P(not C but D owns)
= \((\frac{1}{6} \times \frac{5}{6}) + (\frac{5}{6} \times \frac{1}{6})\)
= \(\frac{5}{36} + \frac{5}{36}\)
= \(\frac{10}{36} = \frac{5}{18}\)
(iii) P(of X, Y and Z, only one owns a car) = P(X owns a car, not Y, not Z) + P(not X, Y owns a car, not Z) + P(not X, not Y, Z owns a car)
= \(3(\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6})\)
= \(3 \times \frac{25}{216}\)
= \(\frac{25}{72}\)