(a) A man P has 5 red, 3 blue and 2 white buses. Another man Q has 3 red, 2 blue and 4 white buses. A bus owned by P is involved in an accident with a bus belonging to Q. Calculate the probability that the two buses are not of the same color.
(b) A man travels from Nigeria to Ghana by air and from Ghana to Liberia by ship. He returns by the same means. He has 6 airlines and 4 shipping lines to choose from. In how many ways can he make his journey without using the same airline or shipping line twice?
P : 5r, 3b, 2w
Q : 3r, 2b, 4w
For Mr P: \(P(r) = \frac{1}{2}; P(b) = \frac{3}{10}; P(w) = \frac{1}{5}\)
For Mr Q: \(P(r) = \frac{1}{3}; P(b) = \frac{2}{9}; P(w) = \frac{4}{9}\)
\(P(\text{ bus of the same color}) = \frac{1}{2} \times \frac{1}{3} + \frac{3}{10} \times \frac{2}{9} + \frac{1}{5} \times \frac{4}{9}\)
= \(\frac{29}{90}\)
\(P(\text{not of the same color}) = 1 - \frac{29}{90} = \frac{61}{90}\)
(b) 6 airlines, 4 shipping lines
If 1 airline is used from Nigeria to Ghana, there will be 5 out of 6 for the return journey. This can be done in \(^{6}C_{5}\) ways. Similarly, if 1 shipping line is used from Ghana to Liberia, there will be 3 shipping lines left for his return journey. This can be done in \(^{4}C_{3}\) ways.
\(\therefore \text{The total number of ways =} ^{6}C_{5} + ^{4}C_{3}\)
= \(6 + 4 = 10\) ways.
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