Originally Posted by

**m777** Hello,

Question no:1

A coin is biased such that a head is thrice as likely to occur as a tail. Find the probability distribution of heads and also find the mean and variance of the distribution when it is tossed 4 times.

(i am confused in this question can somebody explain this.according to point of view"head is thrice "may be we can take as 0.75".

i solve this question by following method it is correct or there is another method plz explain it.)

P(no heads) = 4C0 (.75)0(.25)4 = .004

P(1 head, 3 tails) = 4C1(.75)1(.25)3 = .047

P(2 head, 2 tail) = 4C2(.75)2(.25)2 = .211

P(3 head, 1 tail) = 4C3(.75)3(.25)1 = .423

P(4 head, no tail)= 4C4(.75)4(.25)0 =.316

Mean= E(X) = 0(.004)+1(.047) +2(.211)+3(.423)+4(.316) = 3 heads

Variance = E(X2) -[E(X)]2

E(X2) = 0(.004)+1(.047)+4(.211)+9(.423)+16(.316) = 9.754

Variance = 9.754 - 9 = .754