Here, `n = 6`
P(success)` p = 3/6=1/2 `
P(failure)` q = 3/6 =1/2`
(i)P(5 successes) `= n_(C_5)p^5q^1 = 6_(C_5)(1/2)^5(1/2)`
`=6**1/32**1/2 = 6/64 = 3/32`
(ii)P(At least 5 successes) `= P(X=5)+P(X=6) = 3/32+6_(C_6)(1/2)^6=3/32+1/64 = 7/64`
(iii)P(at most 5 successes) `= 1-P(X=6) = 1-6_(C_6)(1/2)^6 = 1-1/64 = 63/64`