Question

An irregular 6 faced die is thrown And the expectation that in 10 throws it will give five even numbers is twice the expectation that it will give four even number. How many times, 10,000sets of 10 throws would you expect it to give no even number?

Answer 1

Answer:

Let the random variable X denote the number of even numbers.

P(getting x even numbers)

=P(X=x) nCxpxq10−x,x=0,1,2,...,n

Given n=10

∴P(X=x)=10Cxpxq10−x,x=0,1,2,...10

P(getting 5 even numbers) =2P(getting 4 even numbers)

P(X=5)=2P(X=4)=10C5p5q5

=2(10C4p4q6)

4p=3q⇒3p=5q=5(1−p)

⇒8p=5⇒p=85

∴q=1−p=1−85=