(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 own...
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(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.
Answer Details
(a) Two dice give \(6 \times 6 = 36\) equally likely outcomes. A sum of "at least 9" means a sum of 9, 10, 11 or 12.
- Sum 9: (3,6),(4,5),(5,4),(6,3) = 4 ways
- Sum 10: (4,6),(5,5),(6,4) = 3 ways
- Sum 11: (5,6),(6,5) = 2 ways
- Sum 12: (6,6) = 1 way
Favourable outcomes \(= 4+3+2+1 = 10\).
\[ P(\text{sum} \ge 9) = \frac{10}{36} = \frac{5}{18}. \](b) Let \(P(\text{owns a car}) = \frac{1}{6}\), so \(P(\text{does not}) = \frac{5}{6}\). Selections are independent.
(i) Both A and B own a car:
\[ \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}. \](ii) Of C and D, exactly one owns a car (owns-then-not, or not-then-owns):
\[ 2 \times \frac{1}{6} \times \frac{5}{6} = \frac{10}{36} = \frac{5}{18}. \](iii) Of X, Y and Z, exactly one owns a car. Choose which one (\(\binom{3}{1}=3\) ways):
\[ 3 \times \frac{1}{6} \times \left(\frac{5}{6}\right)^{2} = 3 \times \frac{25}{216} = \frac{75}{216} = \frac{25}{72}. \]Download The App On Google Playstore
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