Two binomial probability Researcher paradox

Question

A researcher knows that the probability that a person will respond his letter is 10% If the researcher sends a post-paid letter the probability that the reciever will respond is 40%. The researcher sends 5 post-paid letters and 5 non post-paid letters. What is the probability that he recieves less then 3 responses?

Answer should be : 0,5193 How can you calculate this?

Answer 1

Can I assume this is not homework? So what you have to calculate are the probabilities for 0, 1, and 2 responses. Let's take 0 as the simplest case It will only happen if all five post=paid letters and all five non post-paid letters are answered. By independence it is (0.90)$^5$ (0.60)$^5$. Now add to that the probability that only 1 is returned. There are two ways this can happen, It can be a post-paid returned or a non-postpaid. The disjoint events can have their probabilities summed. For the post-paid case this is (0.60)$^4$ (0.40)$^1$ (0.90)$^5$ But there are 5 ways that 1 post-paid letter can be answered and only 1 way that all five non-post paid letters will not be answered. So this term is 5 (0.60)$^4$ (0.40)$^1$ (0.90)$^5$ and similarly for one non post paid 5 (0.90)$^4$ (0.10) (0.60)$^5$. Last of all you need to add all the cases where 2 letters are answered. This can happen by having 2 non-post-paid letters returned or 1 non-post-paid and 1 post-paid or 2 post-paid. You and the results for these possibilities to the others to get the final answer. The calculations are done in the same way with the number of combinations to get 2 out of 10 letters selected. When they are both from the post-paid group the factor is number of combinations for choosing 2 out of 5 which is 10. The same factor when both are from the non-post-paid group. When it is one from each there are 5 ways for post-paid to match with any one of the 5 non-post-paid. So that factor is 5x5 =25.

Answer 2

If you have R you can get to that by a proper case disjunction. The function dbinom(k, ...) is the probability that you observe exactly $k$ successes and the function pbinom(k, ...) is the probability that you observe $k$ or less successes.

The solution comes out as

dbinom(0, 5, prob=.4)*pbinom(2, 5, prob=.1) +
dbinom(1, 5, prob=.4)*pbinom(1, 5, prob=.1) +
dbinom(2, 5, prob=.4)*pbinom(0, 5, prob=.1)

which is 0.519.