The possible outcomes are:Originally Posted byskhan

hhhh p=0.25^4

thhh p=0.25^3 0.75

hthh p=0.25^3 0.75

hhth p=0.25^3 0.75

hhht p=0.25^3 0.75

tthh p=0.25^2 0.75^2

thth p=0.25^2 0.75^2

thht p=0.25^2 0.75^2

htth p=0.25^2 0.75^2

htht p=0.25^2 0.75^2

hhtt p=0.25^2 0.75^2

ttth p=0.25 0.75^3

ttht p=0.25 0.75^3

thtt p=0.25 0.75^3

httt p=0.25 0.75^3

tttt p=0.75^4

So

P(0 heads)=p(tttt)=0.75^4

P(1 heads)=4 p(httt)=4 0.25 0.75^3

P(2 heads)=6 p(hhtt)=6 0.25^2 0.75^2

P(3 heads)=4 p(hhht)=4 0.25^3 0.75

P(4 heads)=p(hhhh)=0.25^4

This can also be done using the formulae for the binomial distribution.

RonL