If a fair coin is tossed 9 times, find the probability of (a) exactly six heads, (b) at least six heads, (c) at most six heads.

Explanation

The repeated tosses of a coin are Bernoulli trials. Let X denote the no. of heads in an experiment of 9 trials.
Clearly, X has the binomial distribution with n=9, and `p=1/2`.
therefore `P(X=x)=""^nC_x q^(n-x)p^x, " "x=0, 1, 2, .....,n`
here `n=9, p=1/2, q=1-p=1/2`
therefore `P(X=x)=""^9C_x(1/2)^(9-x)(1/2)^x=""^9C_x(1/2)^9`
(i) `P(X=6)=""^9C_6(1/2)^9=(9!)/(6!3!).(1)/2^9=21/128`
(ii) P(at least six heads)=`P(Xgt=6)`
=P(X=6)+P(X=7)+P(X=8)+P(X=9)
`=""^9C_6(1/2)^9+""^9C_7(1/2)^9+""^9C_8(1/2)^9+""^9C_9(1/2)^9`
`=(1/2)^9[(9!)/(6!3!)+(9!)/(7!2!)+(9!)/(8!1!)+(9!)/(9!0!)]`
`=1/512[84+36+9+1]`
`=130/512=65/256`
(iii) P(at most six heads) `=P(Xlt=6)`
=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)
`=(1/2)^9+""^9C_1(1/2)^9+""^9C_2(1/2)^9+""^9C_3(1/2)^9+""^9C_4(1/2)^9+""^9C_5(1/2)^9+""^9C_6(1/2)^9`
`=(1/2)^9[1+9+36+84+126+126+84]`
`=1/2^9[466]=466/512=233/256`