Question

A coin is tossed 4 times. Let X denote the number of heads. Find the probability distribution of X. Also, find the mean and variance of X.

Answer 1

Correct Answer - Mean = 2, variance =1
In a single throw of a coin, P(H) `=(1)/(2) and P(barH)=P(T)=(1)/(2).`
Let X show the number of heads. Then, X=0, 1, 2, 3 or 4.
`P(X=0)=P(" no head")=P(barH_(1) barH_(2) barH_(3) barH_(4))=((1)/(2)xx(1)/(2)xx(1)/(2)xx(1)/(2))=(1)/(16).`
`P(X=1)=P("one head")=P(H_(1) barH_(2) barH_(3) barH_(4)) or (barH_(1) H_(2) barH_(3) barH_(4)) or P(barH_(1) barH_(2) H_(3) barH_(4)) or P(barH_(1) barH_(2) barH_(3) H_(4))`
`=4((1)/(2)xx(1)/(2)xx(1)/(2)x(1)/(2))=(4xx(1)/(6))=(1)/(4).`
`P(X=2)=P("two heads")= P(H_(1) H_(2) barH_(3) barH_(4)) or P(H_(1) barH_(2) H_(3) barH_(4)) or P(barH_(1) H_(2) barH_(3) H_(4)) or P(barH_(1) barH_(2) H_(3) H_(4))`
`=(6xx(1)/(16))=(3)/(8).`
`P(X=3)=P("three heads")=P(H_(1) H_(2) H_(3) barH_(4)) or (H_(1) H_(2) barH_(3) H_(4)) or P(H_(1) barH_(2) H_(3) H_(4)) or P(barH_(1) H_(2) H_(3) H_(3))`
`=4xx((1)/(2)xx(1)/(2)xx(1)/(2)xx(2))=(4xx(1)/(16))=(1)/(4).`
P(X=4) = P(four heads) `=P(H_(1) H_(2) H_(3) H_(4))=((1)/(2)xx(1)/(2)xx(1)/(2)xx(1)/(2))=(1)/(16).`
Thus, we have