please help , probability problem

Question

A box contains 8 hard-centred and 9 soft-centred chocolates. Two are selected at random. What is the probability that one is hard-centred and one is soft-centred ?

Answer 1

There are 17 assorted chocolates. Probability of hard centre=8/17, probability of soft centre=9/17.

 There are 4 ways of selecting 2 chocolates: SS, SH, HS, HH where H=hard and S=soft. Two of these combine H and S.

Suppose the first chocolate is S (probability 9/17), leaving 16 chocolates, which include 8 H; the probability of selecting H next is 8/16.

The combined probability is (9/17)(8/16)=9/34.

Now suppose the first chocolate is H (probability 8/17). Probability of selecting S next is 9/16. Combined probability (8/17)(9/16)=9/34.

So adding these we get 9/34+9/34=9/17.

Probability of picking a hard centre and a soft centre is 9/17.

(Probability of SS=(9/17)(8/16)=9/34; probability of HH=(8/17)(7/16)=7/34. Add these probabilities: 16/34=8/17. When we add all the probabilities together we get 9/17+8/17=1.)