There are 3 green apples, 5 yellow apples, 4 green bananas, and 6 yellow bananas. You are to select two at random. What is the probability of selecting a banana and an apple given that both are yellow?
We have $Pr(\text{apple and banana | both are yellow})=Pr(\text{yellow apple and yellow banana})/Pr(\text{both are yellow})$.
We have 11 yellow apples and yellow bananas, and there are $C(11,2)$ ways to choose two of them. The numerator is $C(11,2)/C(18,2)$. There are also 11 yellow fruits, and there are again $C(11,2)$ ways to choose two yellow fruits. The denominator turns out to be the same. This doesn't look right. Where am I mistaken?
Best Answer
Easy way:
Since it is given that both pieces of fruit are yellow, pretend that the green fruit do not exist.
Probability is
$$\frac{N\text{(umerator)}}{D\text{(enominator)}}$$
where $D = \binom{11}{2}$.
Then $N = \binom{5}{1} \times \binom{6}{1}.$
Explanation:
$D$ represents the total number of ways of selecting two pieces of yellow fruit.
$N$ represents the total number of ways of selecting one yellow apple and one yellow banana.
Addendum
Responding to comment question:
The more formal approach is to use Bayes Theorem.
That is, given events $R,S$ you have that
$p(R|S) = \frac{p(RS)}{p(S)}.$
Let $(R)$ denote the event of 1 yellow apple and 1 yellow banana being chosen.
Let $(S)$ denote the event that two yellow pieces of fruit were chosen.
You want $p(R|S)$.
$p(RS) = [(5/18)\times (6/17)] + [(6/18)\times (5/17)].$
The above expression has two terms, which represents that you can choose either the yellow apple first, or the yellow banana first.
$p(RS)$ simplifies to $\frac{2 \times 5 \times 6}{18 \times 17}.$
$p(S) = 11/18 \times 10/17.$
So $\frac{p(RS)}{p(S)} = \frac{2 \times 5 \times 6}{11 \times 10}.$