A bit string of length 5 is random chosen
E1: the bit string chosen begins with 1
E2: the bit string chosen ends with 1
E3: the bit string chosen has exactly three 1s.
(a) Find p(E1|E3).
(b) Find p(E3|E2).
(c) Find p(E2|E3).
I got the right answer for part a. p(E1|E3) = p(E1 intersecting E3) / p(E3)
for E1 intersecting E3, one spot is already taken up i.e the first spot. This leaves 6 ways of putting the rest of the 1's(2 of them) giving the probability of 6/32. E3 is 5/16. It appears that the answer for a and c is the same.
Im having a hard time understand part b. Can someone explain to me the difference between b and c?
Best Answer
From a computational perspective, the difference is that you divide by $P(E_2)$ in (b) and $P(E_3)$ in (c); that's the only difference.
As far as meaning: if you know that there are exactly three ones, what's the probability that one of them is in the last position? That's the answer to (c); that's why it is $$ \frac{\binom{4}{2}}{\binom{5}{3}}=\frac{6}{10}=\frac{3}{5}, $$ because there are $\binom{5}{3}$ ways to choose which positions get the three ones, and of those $\binom{4}{2}$ involve having a one in the last position.
For (b): you are asked "if there is a one in the last position, what is the probability that there are a total of exactly three ones?". That's the same as the probability that there are exactly two ones in the first four positions.