Question

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

Answer 1

Hint: We see that the event of getting 6 in single throw of die is a Bernoulli’s trial with probability of success $p={\displaystyle \frac{1}{6}}$ and failure $q={\displaystyle \frac{5}{6}}$. We use probability mass function of a random variable $X$ that takes number of as outcomes in binomial distribution $P(X=k)={}^n{C}_k{p}^k{q}^{n-k}={}^n{C}_k{p}^k{(1-p)}^{n-k}$ to find $P(X\le 2)$ to get ${\displaystyle \frac{a}{b}}\cdot {\left({\displaystyle \frac{5}{6}}\right)}^4$.

Complete step-by-step solution:
We know that binomial distribution is a discrete probability distribution which takes number of successes in $n$ Bernoulli’s trials as outcomes with probability of success $p$ and probability failure $q=1-p$.If random variable $X$ follows binomial distribution $\left(X\stackrel{~}{}B(n,p)\right)$ with number of trials $n\in \mathbb{N}$ and probability of success $p\in [0,1]$the probability that we get $k$ successes in $n$ independent trials is given by the probability mass function,
$$P(X=k)={}^n{C}_k{p}^k{q}^{n-k}={}^n{C}_k{p}^k{(1-p)}^{n-k}$$
 We are given the dice thrown 6 times. The probability of getting a 6 from the dice is $p={\displaystyle \frac{1}{6}}$ and not getting 6 is $q={\displaystyle \frac{5}{6}}$. So we have $p+q={\displaystyle \frac{1}{6}}+{\displaystyle \frac{5}{6}}=1$. Hence the event of getting a six in a throw of single die is a Bernoulli’s trial. If we get a 6 in the trial it is success; otherwise failure. $$$$
We assign a random variable $X$ as which takes the number of times we get 6 as outcomes. We are asked to find the probability that we shall get at most 2 sixes which means we get either 0 six, 1 six or 2 sixes. So we have the required probability as
$$\begin{array}{rl}& P(X\le 2)\\ & \Rightarrow P(X=0)+P(X=1)+P(X=2)\end{array}$$
We use the probability mass function for $p={\displaystyle \frac{1}{2}},q={\displaystyle \frac{5}{6}},n=6$ and $k=0,1,2$ respectively in the above step to have;
$$\begin{array}{rl}& \Rightarrow {}^6{C}_0{\left({\displaystyle \frac{1}{6}}\right)}^0{\left({\displaystyle \frac{5}{6}}\right)}^{6-0}+{}^6{C}_1{\left({\displaystyle \frac{1}{6}}\right)}^1{\left({\displaystyle \frac{5}{6}}\right)}^{6-1}+{}^6{C}_2{\left({\displaystyle \frac{1}{6}}\right)}^2{\left({\displaystyle \frac{5}{6}}\right)}^{6-2}\\ & \Rightarrow 1\times 1\cdot \times {\left({\displaystyle \frac{5}{6}}\right)}^6+6\times {\displaystyle \frac{1}{6}}{\left({\displaystyle \frac{5}{6}}\right)}^5+15\times {\displaystyle \frac{1}{36}}{\left({\displaystyle \frac{5}{6}}\right)}^4\\ & \Rightarrow {\left({\displaystyle \frac{5}{6}}\right)}^6+{\left({\displaystyle \frac{5}{6}}\right)}^5+{\displaystyle \frac{5}{12}}{\left({\displaystyle \frac{5}{6}}\right)}^4\end{array}$$
We take ${\left({\displaystyle \frac{5}{6}}\right)}^4$ common to have;
$$\begin{array}{rl}& \Rightarrow {\left({\displaystyle \frac{5}{6}}\right)}^4({\displaystyle \frac{25}{36}}+{\displaystyle \frac{5}{6}}+{\displaystyle \frac{5}{12}})\\ & \Rightarrow {\left({\displaystyle \frac{5}{6}}\right)}^4\left({\displaystyle \frac{25+30+15}{36}}\right)\\ & \Rightarrow {\left({\displaystyle \frac{5}{6}}\right)}^4\left({\displaystyle \frac{70}{36}}\right)\end{array}$$
We are given that the probability is ${\displaystyle \frac{a}{b}}\cdot {\left({\displaystyle \frac{5}{6}}\right)}^4$. So we have the required answer as
$$\begin{array}{rl}& {\left({\displaystyle \frac{5}{6}}\right)}^4\left({\displaystyle \frac{70}{36}}\right)={\displaystyle \frac{a}{b}}\cdot {\left({\displaystyle \frac{5}{6}}\right)}^4\\ & \Rightarrow {\displaystyle \frac{a}{b}}={\displaystyle \frac{70}{36}}\\ & \Rightarrow a=70,b=36\\ & \Rightarrow a+b=70+36=106\end{array}$$

Note: We note that we are not given in the question that ${\displaystyle \frac{a}{b}}$ is in simplest form otherwise ${\displaystyle \frac{a}{b}}={\displaystyle \frac{35}{18}}$. We also note that a Bernoulli trial is a random experiment with exactly two possible outcomes called success or failure with probability of success does not change by repeating the experiment. Here in this problem we can either get 6 in a single throw of die or not get it at all. The expectation of binomial distribution is $np$ and the variance is $np(1-p)$.