Question

The probability of throwing at most 2 sixes in 6 throws of a single die is $\dfrac{a}{b}\cdot {{\left( \dfrac{5}{6} \right)}^{4}}$. 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 $p=\dfrac{1}{6}$ and failure $q=\dfrac{5}{6}$. We use probability mass function of a random variable $X$ that takes number of as outcomes in binomial distribution $P\left( X=k \right)={}^{n}{{C}_{k}}{{p}^{k}}{{q}^{n-k}}={}^{n}{{C}_{k}}{{p}^{k}}{{\left( 1-p \right)}^{n-k}}$ to find $P\left( X\le 2 \right)$ to get $\dfrac{a}{b}\cdot {{\left( \dfrac{5}{6} \right)}^{4}}$.

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 $\left( X\tilde{\ }B\left( n,p \right) \right)$ with number of trials $n\in \mathsf{\mathbb{N}}$ and probability of success $p\in \left[ 0,1 \right]$the probability that we get $k$ successes in $n$ independent trials is given by the probability mass function,
\[P\left( X=k \right)={}^{n}{{C}_{k}}{{p}^{k}}{{q}^{n-k}}={}^{n}{{C}_{k}}{{p}^{k}}{{\left( 1-p \right)}^{n-k}}\]
 We are given the dice thrown 6 times. The probability of getting a 6 from the dice is $p=\dfrac{1}{6}$ and not getting 6 is $q=\dfrac{5}{6}$. So we have $p+q=\dfrac{1}{6}+\dfrac{5}{6}=1$. 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
\[\begin{align}
  & P\left( X\le 2 \right) \\
 & \Rightarrow P\left( X=0 \right)+P\left( X=1 \right)+P\left( X=2 \right) \\
\end{align}\]
We use the probability mass function for $p=\dfrac{1}{2},q=\dfrac{5}{6},n=6$ and $k=0,1,2$ respectively in the above step to have;
\[\begin{align}
  & \Rightarrow {}^{6}{{C}_{0}}{{\left( \dfrac{1}{6} \right)}^{0}}{{\left( \dfrac{5}{6} \right)}^{6-0}}+{}^{6}{{C}_{1}}{{\left( \dfrac{1}{6} \right)}^{1}}{{\left( \dfrac{5}{6} \right)}^{6-1}}+{}^{6}{{C}_{2}}{{\left( \dfrac{1}{6} \right)}^{2}}{{\left( \dfrac{5}{6} \right)}^{6-2}} \\
 & \Rightarrow 1\times 1\cdot \times {{\left( \dfrac{5}{6} \right)}^{6}}+6\times \dfrac{1}{6}{{\left( \dfrac{5}{6} \right)}^{5}}+15\times \dfrac{1}{36}{{\left( \dfrac{5}{6} \right)}^{4}} \\
 & \Rightarrow {{\left( \dfrac{5}{6} \right)}^{6}}+{{\left( \dfrac{5}{6} \right)}^{5}}+\dfrac{5}{12}{{\left( \dfrac{5}{6} \right)}^{4}} \\
\end{align}\]
We take ${{\left( \dfrac{5}{6} \right)}^{4}}$ common to have;
\[\begin{align}
  & \Rightarrow {{\left( \dfrac{5}{6} \right)}^{4}}\left( \dfrac{25}{36}+\dfrac{5}{6}+\dfrac{5}{12} \right) \\
 & \Rightarrow {{\left( \dfrac{5}{6} \right)}^{4}}\left( \dfrac{25+30+15}{36} \right) \\
 & \Rightarrow {{\left( \dfrac{5}{6} \right)}^{4}}\left( \dfrac{70}{36} \right) \\
\end{align}\]
We are given that the probability is $\dfrac{a}{b}\cdot {{\left( \dfrac{5}{6} \right)}^{4}}$. So we have the required answer as
\[\begin{align}
  & {{\left( \dfrac{5}{6} \right)}^{4}}\left( \dfrac{70}{36} \right)=\dfrac{a}{b}\cdot {{\left( \dfrac{5}{6} \right)}^{4}} \\
 & \Rightarrow \dfrac{a}{b}=\dfrac{70}{36} \\
 & \Rightarrow a=70,b=36 \\
 & \Rightarrow a+b=70+36=106 \\
\end{align}\]

Note: We note that we are not given in the question that $\dfrac{a}{b}$ is in simplest form otherwise $\dfrac{a}{b}=\dfrac{35}{18}$. 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 $np\left( 1-p \right)$.