4) (20 pts) A box contains 5 apples and 6 oranges. Four children each receive a fruit from the box, one after the other, randomly chosen, without replacement. What is the probability that all four children receive the same fruit

Probability of an event = Number of outcomes favourable to that event/ Number of all possible outcomes.

The set of all possible outcomes is called the sample space and is denoted by S.

Any 3 fruits out of 11 in the box can be randomly selected in 11C3= (11x10 x 9)/(1 x 2x3) = 165 ways. Hence n(S) = 165.

We want the desired event E that the 1 fruit of each type is chosen. Now 1 apple out of 5 can be chosen in 5C1 = 5 ways, 1 out of 4 oranges in 4C1 = 4 and 1 banana out of 2 in 2C1 = 2 ways. Therefore, by fundamental counting principle, one of each type of fruit can be chosen in 5x4x2 = 40 ways So n(E) = 40.

Hence the probability of getting three fruits one of each type is = n(E)/ n(S) = 40/165 = 8/33.

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