Probablity_expectation_doubt

Question

A coin is biased so that the probability a head comes up when it is flipped is 0.6. What is the expected number of heads that come up when it is flipped 10 times? My approach and doubt-: can't i use bernouli trials formula(n*p) n=10 p=0.6 is 6 the answer ? if not then what is the correct answer?

Answer 1

Let $X$ denote the number of heads, then:

• $P(X=n)=\binom{10}{n}\cdot(0.6)^{n}\cdot(1-0.6)^{10-n}$
• $E(X)=\sum\limits_{n=0}^{10}n\cdot\binom{10}{n}\cdot(0.6)^{n}\cdot(1-0.6)^{10-n}=6$