12a. 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 the same. Leave your answer in this format ' a/b.'
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?
12a. Total buses owned by P: 10
Total buses owned by Q: 9
Total possible pairs of buses (one from P and one from Q): \( 10 \times 9 = 90 \)
Pairs of the "same colour":
Both red: \( 5 \times 3 = 15 \)
Both blue: \( 3 \times 2 = 6 \)
Both white: \( 2 \times 4 = 8 \)
Total same colour: 15 + 6 + 8 = 29
Pairs of "different colours": 90 - 29 = 61
Probability that the two buses are not the same colour: \(\frac{61}{90}\)
12b. The journey has four independent legs:
Nigeria → Ghana by air: 6 airline choices
Ghana → Liberia by ship: 4 shipping line choices
Liberia → Ghana by ship (return): 4 shipping line choices (same line can be used again)
Ghana → Nigeria by air (return): 6 airline choices (same airline can be used again)
Since repetition of the same airline (for the two air legs) and the same shipping line (for the two ship legs) is allowed, the total number of ways is:
= 6 × 4 × 4 × 6 = 576 ways
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