Question

(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?​

Answer 1

(a) To calculate the probability that the two buses involved in the accident are not of the same color, we can consider the total number of possible outcomes and the favorable outcomes where the buses are of different colors.

Total number of outcomes = Total number of buses owned by P * Total number of buses owned by Q = (5 + 3 + 2) * (3 + 2 + 4) = 10 * 9 = 90

Favorable outcomes = Number of ways to choose a bus of different colors from P's buses * Number of ways to choose a bus of different colors from Q's buses = (5 * 9) + (3 * 7) + (2 * 7) = 45 + 21 + 14 = 80

Therefore, the probability that the two buses involved in the accident are not of the same color is 80/90, which simplifies to 8/9.

(b) The man has 6 airlines and 4 shipping lines to choose from for each leg of his journey.

For the first leg (from Nigeria to Ghana), he can choose any of the 6 airlines.

For the second leg (from Ghana to Liberia), he can choose any of the 4 shipping lines.

For the return journey, he has the same options as the first leg.

Since the choices for each leg are independent, the total number of ways he can make his journey without using the same airline or shipping line twice is:

Number of ways = Number of choices for the first leg * Number of choices for the second leg * Number of choices for the return journey

= 6 * 4 * 6

= 144

Therefore, the man can make his journey without using the same airline or shipping line twice in 144 different ways.