A coin is tossed 6 times. Find the probability of getting
(i) exactly 4 heads (ii) at least 1 head (iii) at most 4 heads .

Answer

`(i)(15)/(64) (ii) .(63)/(64) (iii) .(57)/(64)`

Explanation

P(a head ) `=(1)/(2) ` and P (not a head ) `=(1)/(2)`
`:. , p =(1)/(2) , q= (1)/(2) " and " n=6`
`:. P(X=r) =.^(n)C_(r ) .p^(r ) .q^((n-r)) =.^(6)C_(r ) .((1)/(2))^(6)`
(i) P (exactly 4 heads ) `=.^(6)C_( 4) .((1)/(2))^(6)`
(ii) P (at least 1 head) `=1- P` (no head)
=1- P( 0 head) `=[1-,^(6)C_(0).((1)/(2))^(6)]`
(iii) P (at the most 4 heads ) =P ( 4 or less than 4 heads )
=1- P[5 heads or 6 heads ]
`=1 -[(.^(6)C_(5)+.^(6)C_(6)).((1)/(2))^(6)]`