F.3 Probability問題

P(draw two apples) = (2/10) (1/9) = 2/90
[there were 2 apples out of 10 . after you drawn an apple, remain 1 apple out of 9]

P(draw two pears) = (3/10) (2/9) = 6/90

P(draw two oranges) = (5/10) (4/9) = 20/90

P(draw two fruits of the same kind)
= P(draw two apples or draw two pears or draw two oranges)
= P(draw two apples) + P(draw two pears) + P(draw two oranges)
= 2/90 + 6/90 + 20/90
= 28/90

P(draw two fruits of different kind)
= 1 - P(draw two fruits of the same kind)
= 1 - 28/90
= 62/90
= 31/45

A bag contains 2 apples, 3 pears and 5 oranges.
If Mary draws two fruit at random, one by one without replacement.
Assume that all theapples, pears and oranges are distinguishable.
What is the probability that she draw two different kinds of fruit?There are 10 fruit and (10C2) = 45 way to draw two fruit.
If she draw two different kinds of fruit,
it may be apples and pears (2*3 ways), pears and oranges (3*5), oranges and apples (5*2)
So she has 2*3 + 3*5 + 5*2 = 31 ways to draw two different kinds of fruit.
Hence the probability is 31/45.