Question

A die is thrown $6$ times. If "getting an odd number" is a success, what is the probability of
(i) $5$ success?
(ii) at least $5$ successes ?

Answer 1

We know that,

the number of dice $=6$

$n\left(s\right)=6$

According to given condition $=p(1,3,5)$

$p\left(\epsilon \right)=p=\frac{3}{6}=\frac{1}{2}$

now we know that

$p\left(\epsilon \right)+p\left(\epsilon \right)=1$

let $p\left(\epsilon \right)=p$ and $p\left(\epsilon \right)=q$

then, $p+q=1$

$q=1-p=1-\frac{1}{2}=\frac{1}{2}$

$q=\frac{1}{2}$

$\left(i\right)5successes=p\left(5\right){=}^6{C}_5{\left(\frac{1}{2}\right)}^{6-5}{\left(\frac{1}{2}\right)}^5$

$=6\times {\left(\frac{1}{2}\right)}^6=6\times \frac{1}{64}=\frac{3}{32}$

$p\left(5\right)=\frac{3}{32}$

$\left(ii\right)$at least $5$ successes $=p\left(5\right)+p\left(6\right)$

${=}^6{C}_5\left(\frac{1}{2}\right){\left(\frac{1}{2}\right)}^5{+}^6{C}_6{\left(\frac{1}{2}\right)}^{6-6}{\left(\frac{1}{2}\right)}^6$

$=\left[{}^6{C}_5{+}^6{C}_6\right]{\left(\frac{1}{2}\right)}^6$

$=(6+1)\frac{1}{64}{\therefore }^6{C}_5=6$

${}^6{C}_6=1$

$p\left(5\right)+p\left(6\right)=\frac{7}{64}$

Hence, this is the answer.