Answer:
Step-by-step explanation:
Given that a random customer buys as follows
A B C Nothing
Prob 0.25 0.30 0.35 0.10
a) Prob for atleast one customer out of 4 buying one = 1-Prob that all 4 do not buy = [tex]1- 0.1^4 =0.9999[/tex]
(since independent)
b) the probability that the 3rd purchase of brand B is made before the 6th customer makes his purchase
=prob that out of 6 customers atleast one purchase brand 3
= P(x>=1) where X = no of customers buying brand B and n = 6
Here X is binomial: p = 0.30 and q = 0.70
Hence P(X>=1) = [tex]1-q^6[/tex]
=[tex]1-0.7^6\\=0.8823[/tex]