Question

The probability of throwing at most $2$ sixes in $6$ throws of a single die is $=\frac{a}{b}\cdot {\left(\frac{5}{6}\right)}^4$. Find $a+b$

Answer 1

The repeated tossing of the die are Bernoulli trials. Let $X$ represent the number of times of getting sixes in 6 throws of the die.
Probability of getting six in a single throw of die, $p=\frac{1}{6}$
$\therefore q=1-p=1-\frac{1}{6}=\frac{5}{6}$
Clearly, $X$ has a binomial distribution with $n=6$
$\therefore P(X=x){=}^n{C}_x{q}^{n-x}{p}^x{=}^6{C}_x{\left(\frac{5}{6}\right)}^{6-x}\cdot {\left(\frac{1}{6}\right)}^x$
$P\left(atmost2sixes\right)=P(X\le 2)$
$=P(X=0)+P(X=1)+P(X=2)$
${=}^6{C}_0{\left(\frac{5}{6}\right)}^6{+}^6{C}_1{\left(\frac{5}{6}\right)}^5\cdot \left(\frac{1}{6}\right){+}^6{C}_2{\left(\frac{5}{6}\right)}^4\cdot {\left(\frac{1}{6}\right)}^2$
$=1\cdot {\left(\frac{5}{6}\right)}^6+6\cdot \frac{1}{6}\cdot {\left(\frac{5}{6}\right)}^5+15\cdot \frac{1}{36}\cdot {\left(\frac{5}{6}\right)}^4$
$={\left(\frac{5}{6}\right)}^6+{\left(\frac{5}{6}\right)}^5+\frac{5}{12}\cdot {\left(\frac{5}{6}\right)}^4$
$={\left(\frac{5}{6}\right)}^4[{\left(\frac{5}{6}\right)}^2+\left(\frac{5}{6}\right)+\left(\frac{5}{12}\right)]$
$={\left(\frac{5}{6}\right)}^4\cdot \left[\frac{25}{36}+\frac{5}{6}+\frac{5}{12}\right]$
$={\left(\frac{5}{6}\right)}^4\cdot \left[\frac{25+30+15}{36}\right]$
$=\frac{70}{36}\cdot {\left(\frac{5}{6}\right)}^4$
$=\frac{35}{18}\cdot {\left(\frac{5}{6}\right)}^4$