Six coin are tossed simultaneously . Find the probability of getting
(i) 3 heads (ii) no head (iii) at least one head
(iv) not more than 3 heads .

Explanation

The experiment may be be taken as throwing a single coin 6 times .
In a single throw of a coin , we have S={H,T}
P(getting a head) `=(1)/(2)` and P(not getting a head) `=(1-(1)/(2)) =(1)/(2)`
Let X be the random varibale showing the number of heads
`P(X=r)=.^(n)C_(r ).p^(r ) .q^((n-r)) =.^(6)C_(r ) .((1)/(2))^(r ) .((1)/(2))^(6-r) =.^(6)C_(r ) .((1)/(2))^(6)`
(i) P(getting 3 heads) =P(X=)3 `=.^(6)C_(3) .((1)/(2))^(6)=(5)/(16).`
(ii) P(getting no head) `=P(X =0)= .^(6)C_(0).((1)/(2))^(6) =(1)/(64)`
(iii) P(getting at least 1 head)
`=1- P(X =0) =1- [.^(6)C_(0).((1)/(2))^(6)]`
`=(1-(1)/(64)) =(63)/(64)`
(iv) P(getting not more than 3 heads)
=P(no head or 1 head or 2 heads or 3 heads)
`=P(X=0) +P(X=1)+P(X=2)+P(X=3)`
`=.^(6)C_(0) .((1)/(2))^(6)+.^(6)C_(1).((1)/(2))^(6) +.^(6)C_(2).((1)/(6))^(6)+.^(6)C_(3).((1)/(2))^(6)`
`=((1)/(2))^(6) .(1+6+15+20)=((1)/(64) xx 42) =(21)/(32)`