Question

A die 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 ?

Answer 1

The repeated tosses of a die are Bernoulli trials. Let $X$ denote the number of successes of getting odd numbers in an experiment of 6 trials.
Probability of getting an odd number in a single throw of a die is, $p=\frac{3}{6}=\frac{1}{2}$
$\therefore q=1-p=\frac{1}{2}$
$X$ has a binomial distribution.
Therefore, $P(X=x){=}^n{C}_{n-x}{q}^{n-x}{p}^x$, where $x=0,1,\mathrm{2....}n$
${=}^6{C}_x{\left(\frac{1}{2}\right)}^{6-x}\cdot {\left(\frac{1}{2}\right)}^x$
(i) $P\left(5successes\right)=P(X=5)$
${=}^6{C}_5{\left(\frac{1}{2}\right)}^6$
$=6\cdot \frac{1}{64}$
$=\frac{3}{32}$

(ii)$P\left(atleast5successes\right)=P(X\ge 5)$
$=P(X=5)+P(X=6)$
${=}^6{C}_5{\left(\frac{1}{2}\right)}^6{+}^6{C}_6{\left(\frac{1}{2}\right)}^6$
$=6\cdot \frac{1}{64}+1\cdot \frac{1}{64}$
$=\frac{7}{64}$

(iii) $P\left(atmost5successes\right)=P(X\le 5)$
$=1-P(X>5)$
$=1-P(X=6)$
$=1{-}^6{C}_6{\left(\frac{1}{2}\right)}^6$
$=1-\frac{1}{64}$
$=\frac{63}{64}$