# A die is thrown 6 times. If “getting an odd number” is a “success”, what is the probability of (i) 5 successes ? (ii) atmost 5 successes ?

Given $n=6$

Success = Getting an odd number  $=\left \{ 1,3,5 \right \}$

p : probablity of success

q : probablity of failure

$p=\frac{3}{6}=\frac{1}{2}$,  $q=\frac{3}{6}=\frac{1}{2}$

(i)  $P(5\: \: successes)=\!^{n}\textrm{C}_5p^5q^1$

$=\! ^6\textrm{C}_{5} \left ( \frac{1}{2} \right )^5\! \left ( \frac{1}{2} \right )^1$

$=6 \times \frac{1}{32}\times \frac{1}{2}=\frac{6}{64}=\frac{3}{32}$

(ii) $P(atmost\: \: successes)=P(x\leq 5)$

$=1-P(x=6)$

$=1-\!^{6}\textrm{C}_6 \left ( \frac{1}{2} \right )^6$

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

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