A dice is thrown 6 times. If getting an odd number is a success, what is the probability of
(i) 5 successes ?

(ii) at least 5 successes ?

(iii) at most 5 successes ?

Probability of getting an odd number in one trial
$= \cfrac{3}{6} = \cfrac{1}{2} = p$ (say)

Probability of getting an even number in one trial
$= \cfrac{3}{6} = \cfrac{1}{2} = q$ (say)

A has a binomial distribution with $n = 6,p = \cfrac{1}{2},q = \cfrac{1}{2}$

$P(X = r){ = ^n}{C_r}{q^{n - r}} - {p^r}$

(i) $P(5{\rm{ successes }}) = P(5)$

${ = ^6}{C_5}q{p^5} = (6)\left( {\cfrac{1}{2}} \right){\left( {\cfrac{1}{2}} \right)^5} = 6\left( {\cfrac{1}{{64}}} \right) = \cfrac{3}{{32}}$

(ii) $P$ (at least 5 successes) $= P(X = 5) + P(X = 6)$

$= \cfrac{3}{{32}}{ + ^6}{C_6}{q^n}{p^6}$

(Using Part (i))
$= \cfrac{3}{{32}} + (1)(1){\left( {\cfrac{1}{2}} \right)^6} = \cfrac{3}{{32}} + \cfrac{1}{{64}} = \cfrac{7}{{64}}$

(iii) $P$ (at most 5 successes)
$= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$ $+ P(X = 5) = 1 - P(6)$

$= 1 - \cfrac{1}{{64}} = \cfrac{{63}}{{64}}$

(Using Part (ii)