Question

Five dice are thrown simultaneously. If the occurrence of an even number in a single die is considered a success and the probability of at most 3 success is $\dfrac{13}{{{2}^{x}}}$, then the value of $x$ is ___.

Answer 1

Hint: We find the probability of getting an even number $p$ and odd number $q$. We see that the random variable that takes the number of successes as outcomes follows binomial distribution because we get $p+q=1$. We find the probability of at most 3 success $P\left( X\le 3 \right)=1-P\left( X>3 \right)=1-\left\{ P\left( X=4 \right)+P\left( X=5 \right) \right\}$ using the formula for the probability of getting $k$ successes in $n$ trials $P\left( X=k \right)={}^{n}{{C}_{k}}{{p}^{k}}{{q}^{n-k}}={}^{n}{{C}_{k}}{{p}^{k}}{{\left( 1-p \right)}^{n-k}}$.

Complete step-by-step solution:
We know that the binomial distribution with parameters $n\in N$ and $p\in \left[ 0,1 \right]$ is the discrete probability distribution of the number of successes in a sequence of $n$ independent experiments (also called trials ) each asking a yes-no question, and each with its own Boolean-valued outcome: success/ yes/true/one (with probability $p$) or failure/no/false/zero (with probability q = 1 − p). \[\]
We also know that random variable $X$ that follows binomial distribution then the probability of getting $k$ successes in $n$ trials is given by,
\[P\left( X=k \right)={}^{n}{{C}_{k}}{{p}^{k}}{{q}^{n-k}}={}^{n}{{C}_{k}}{{p}^{k}}{{\left( 1-p \right)}^{n-k}}\]
We are given in the question that five dice are thrown simultaneously and if the occurrence of an even number in a single die is considered a success and the probability of at most 3 success is $\dfrac{13}{{{2}^{x}}}$ . The number of possible outcomes of a single dice throw is 6 (1, 2, 3, 4, 5, and 6). The number of even numbers is 3(2, 4, 6) and the number of odd numbers is 3 (1, 3, and 5). Let us denote the probability of getting an even number be $p$ and the probability of getting an odd number as $q$. So we have
\[\begin{align}
  & p=\dfrac{3}{6}=\dfrac{1}{2} \\
 & q=\dfrac{3}{6}=\dfrac{1}{2} \\
 & p+q=\dfrac{1}{2}+\dfrac{1}{2}=1 \\
 & \Rightarrow q=1-p \\
\end{align}\]
We can either get an odd or even number not both. So the random variable that counts getting an even number as a success follows binomial distribution. Let the random variable be $X$. We have from the question the probability of at most 3 successes is $\dfrac{13}{{{2}^{x}}}$ . So we have,
\[P\left( X\le 3 \right)=\dfrac{13}{{{2}^{x}}}.........\left( 1 \right)\]
We use the fact that the sum of probabilities when a random variable takes every outcome is 1 and have,
\[P\left( X\le 3 \right)=1-P\left( X>3 \right)\]
There are maximum 5 trials. So the number of successes greater than 3 can be either 4 or 5. So we have,
\[\Rightarrow P\left( X\le 3 \right)=1-\left( P\left( X=4 \right)+P\left( X=5 \right) \right)\]
We use the formula for probability of $k=4,5$ successes in $n=5$ trials in Binomial distribution with probability of success $p=\dfrac{1}{2}$ and have,
\[\begin{align}
& \Rightarrow P\left( X\le 3 \right)=1-\left( {}^{5}{{C}_{4}}{{\left( \dfrac{1}{2} \right)}^{4}}{{\left( 1-\dfrac{1}{2} \right)}^{5-4}}+{}^{5}{{C}_{4}}{{\left( \dfrac{1}{2} \right)}^{5}}{{\left( 1-\dfrac{1}{2} \right)}^{5-5}} \right) \\
& \Rightarrow P\left( X\le 3 \right)=1-\left( {}^{5}{{C}_{4}}{{\left( \dfrac{1}{2} \right)}^{5}}+{}^{5}{{C}_{5}}{{\left( \dfrac{1}{2} \right)}^{5}} \right) \\
& \Rightarrow P\left( X\le 3 \right)=1-\left( 5\times \dfrac{1}{32}+1\times \dfrac{1}{32} \right) \\
&\Rightarrow P\left( X\le 3 \right)=1-\dfrac{6}{32}=\dfrac{26}{32}=\dfrac{13}{16}=\dfrac{13}{{{2}^{4}}}....\left( 2 \right) \\
\end{align}\]
We compare right hand sides of equation (1) and (2) to have the required value as,
\[\begin{align}
  & \dfrac{13}{{{2}^{x}}}=\dfrac{13}{{{2}^{4}}} \\
 & \Rightarrow {{2}^{x}}={{2}^{4}} \\
 & \Rightarrow x=4 \\
\end{align}\]

Note: We need to be careful that at most 3 successes are denoted as $P\left( X\le 3 \right)$ not $P\left( X<3 \right)$ or $P\left( X=3 \right)$. We note that a single trial is in binomial distribution called Bernoulli’s trial and for $n=1$ binomial distribution is called Bernoulli’s distribution. The man of binomial distribution is $np$ and the variance of binomial distribution is $np\left( 1-p \right)$. We must be careful of the confusion between binomial and Poisson distribution where the number of trials is infinite.