How express numerator of: Given that after 10 days Ian has had 8 invitations, find the P that he will receive the 9th invitation on the 14th day?

Question

Below is the question that I have been working with and I am stuck on part (c). I approached this question by working with conditional probability, and hence the formula P(A given B)=P(A and B)/P(B), where A represents eight invitations in 10 days and B that he joins the group on the fourteenth day. For the numerator, I had: (10C8 * 0.8^8 * 0.2^2) * (3C0 * 0.8^0 * 0.2^3) * 0.8, ie. the probability of eight invitations in ten days times zero invitations in the next three days and one invitation on the last day, the fourteenth day. And for the denominator, I had 0.0553 (as in the key). The answer to this is 0.0350, however, the key says 0.0064.

What am I doing wrong and what is the correct expression for the numerator and why?

Answer 1

It's probably easier to use the independence property, rather than the definition of conditional probability. If he already had 8 invitations after 10 days, and he is first able to join the group on day 14, that means he must not have gotten an invitation on days 11, 12, or 13, and then got one on day 14. This has probability $0.2^3 * 0.8 = 0.0064$. We don't need to take into account the precise sequence by which he had gotten the 8 invitations in the previous 10 days, as we only care about what is happening on days 11-14 for the purposes of part (c).

For the numerator, I had: (10C8 * 0.8^8 * 0.2^2) * (3C0 * 0.8^0 * 0.2^3) * 0.8, ie. the probability of eight invitations in ten days times zero invitations in the next three days and one invitation on the last day, the fourteenth day. And for the denominator, I had 0.0553 (as in the key). The answer to this is 0.0350, however, the key says 0.0064.

Your denominator should also have been $\binom{10}{8} * 0.8^8 * 0.2^2$, which cancels, leaving only $\binom{3}{0} * 0.8^0 * 0.2^3 * 0.8 = 0.2^3 * 0.8 = 0.0064$, as in the answer key. Your different answer was most likely the result of a computation error.

Answer 2

For part c, he needs to NOT get an invitation on days 11, 12, & 13, and he must get an invitation on day 14. These are all independent events, so you can just multiply the probabilities together: $(1-0.8)^3\cdot(0.8)^1=.0064$

Answer 3

Let $N_{k}=\sum_{i=1}^k X_i$ be the number of invitations by day $k$ where $X_i$ are i.i.d bernoulli random variables with probability of success $0.8$. Then we want $$ P(N_{14}=9, N_{13}\neq 9, N_{12}\neq 9, N_{11}\ne 9\mid N_{10}=8)=\frac{P(N_{10}=8, X_{11}=0, X_{12}=0, X_{13}=0, X_{14}=1)}{P(N_{10}=8)}\tag{0} $$ but by independence it follows that $$ P(N_{10}=8, X_{11}=0, X_{12}=0, X_{13}=0, X_{14}=1)=P(N_{10}=8)P( X_{11}=0, X_{12}=0, X_{13}=0, X_{14}=1) $$ Hence the desired probability $(0)$ is given by $$ P( X_{11}=0, X_{12}=0, X_{13}=0, X_{14}=1)=(0.2)^3\times0.8 $$ where we again use independence.