The probability that a patient recovered from a rare blood disease is 0.4.If 15 people are known to have contracted this disease,

Question

What is the brobability that

a.at least 10 survived

b.from 3 to 8 survived

c.exactly 5 survived

Answer 1

We need to expand the binary (0.4+0.6)^15. This will give us discreet probabilities for each eventuality from no survivors up to all survivors.

The coefficients for the expansion are 1, 15, 105, 455, 1365, 3003, 5005, 6435, 6435, 5005, 3003, 1365, 455, 105, 15, 1 corresponding to all survivors, 14 survivors, 13 survivors, ..., no survivors.

If k is the number of survivors to be calculated then the relevant coefficient is applied to the product 0.4^k*0.6^(15-k).

a) At least 10 survivors is the sum of the probabilities for k ranging from 15 to 10, the sum of the first 6 terms=0.0338 approx or 3.38%. CHECKED

b) We need the sum from k=8 to 3=0.8778 or 87.78%. CHECKED

c) Exactly 5 is 3003*0.4^5*0.6^10=0.1859 or 18.59%. CHECKED

Tables for binomial expansion should give the same results.