Two different methods but not same answer

Question

I wanted to work this problem 2 different ways, but am getting 2 different answers. Can you assist in helping me understand why method 2 doesn't work?

Note: There are 16 letters and 2 E's and 3 B's.

Method 1: Prob of getting an E on the first draw + Prob getting a B on the second draw - minus the prob of getting EC (overlap).

$2/16 + 3/15 - (2/16 \cdot3/15) =3/10$ correct answer!

Method 2: Prob of getting and E on the first draw and NOT a B on the second, PLUS Prob of getting NOT E on the first draw and a B on the second, PLUS Prob of getting an EB. Add together all the possibilities.

Prob of E on first and NOT B on second = $2/16 \cdot13/15=26/240$
Prob of NOT E on first and B on second = $14/16 \cdot3/15 = 42/240$
Prob of E on first and B on Second = $2/16 \cdot3/15 = 6/240 $

This Method 2 does not produce the $3/10$ answer. I don't understand why not.

Answer 1

The probability of E on first and not B on second is $\frac{2}{16}*\frac{\color{red}{12}}{15}$

So adding them all up you do once again get $\frac{72}{240}=\frac{3}{10}$

BUT:

The answer $\frac{3}{10}$ is NOT correct!

The probability of getting a B on the second draw depends on whether you get a B on the first draw or not: if you get a B on the first draw, then the probability of getting a B on the second is only $\frac{2}{15}$.

So, the probability of getting a B on the second draw is $P(B \: first)*\frac{2}{15}+P(not \: B \: first)*\frac{3}{15}=\frac{3}{16}*\frac{2}{15}+\frac{13}{16}*\frac{3}{15} = \frac{45}{240}$

Hence, using your first method, you get that E on first or B on second is $\frac{2}{16} + \frac{45}{240} - \frac{6}{240} = \frac{69}{240}=\frac{23}{80}$

Using your second method, you get that the probability of not gettin a E first but getting second needs to be changed as well, because the not E on first means you can get a B first .. Or something other than E or B first.

So, the probability of getting not E first and B second is $\frac{3}{16}*\frac{2}{15}+\frac{11}{16}*\frac{3}{15}=\frac{39}{240}$

So adding all three probabilities in the second method you get $\frac{24}{240}+\frac{39}{240}+\frac{6}{240}=\frac{69}{240}=\frac{23}{80}$

Answer 2

If the first letter is not an E, it may be a B, in which case in the second draw the probability of drawing a B is reduced to $2/15$. So, one needs to distinguish three disjoint events that amount to success:

• The first letter drawn is an E.
• The two letters are both Bs.
• The first letter is neither E nor B, and the second letter is a B.

Adding the probabilities, we get:

$$ \frac{2}{16} + \frac{3}{16} \cdot \frac{2}{15} + \frac{11}{16}\cdot \frac{3}{15} = \frac{23}{80} \enspace. $$