sk8ternite wrote:A basket contains 5 apples, of which 1 is spoiled and the rest are good. If Henry is to select 2 apples from the basket simultaneously and at random, what is the probability that the 2 apples selected will include the spoiled apple?
(a) 1/5
(b) 3/10
(c) 2/5
(d) 1/2
(e) 3/5

Answer is c.

I am trying to figure out why you add the 2 scenarioes instead of multiplying.

4C1 / 5C2 = 2 / 5

Can you provide a little more detail to your explanation?

the way I thought about it was:

the probability of picking the bad apple first is 1/5, and then the 2nd pick would be 1, so scenario would is 1/5*1=1/5.

But you could pick the bad apple 2nd, so it would be: 4/5*1/5=4/20.

If either situation can occur, it is an "or" situation and you add the two scenarios. 1/5+4/20= 8/20 or 2/5

Is this correct? Could someone also explain the combinatorics way of doing the problem?

probabilty of not selecting the spoiled one:
4C2/5C2= 3/5
so prob of selecting the spoiled one=1-3/5=2/5

4 GOOD APPLES AND 1 BAD APPLE...

No of ways for chosing 1 bad apple is 1C1.
No of ways for chosing 1 Good Apple is 4C1.

Therefore, no of ways for chosing the apples ( as given under problem )are ( 1C1*4C1) = 4.

Total number of ways for chosing 2 apples from 5 apples is ( Denominator part ) = 5C2 = 10.

Therefore the required probability is (4/10) = (2/5).

Hope it is clear enough..