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. Select the correct answer from above options

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