(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?
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The b) part of this question is not correct.
Going from Nigeria to Ghana, he has an option of taking any one of 6 airlines. From Ghana to Liberia, he can chose any one of 4 shipping airline. On his way, back, since he can't enter the previous airline or shipping line, then he can go back from Liberia to Ghana by choosing one of the remaining 3 shipping lines. From Ghana to Lagos, he can enter any one of 5 airlines.
Therefore, the total way he can complete the journey is simple, 6×4×3×5=360
The solution of b) should be
6P2 x 4P2 = 360.
This is because you are arranging 2 items out of 6 and another 2 items out of 4.
Assuming the airlines are ABCDE and F.
The possible ways are AB, BA ... meaning that the arrangement matters and not just selection.
Also, you cannot use + as it is a conjunction AND not a Disjunction OR.
