Question

An experiment succeeds twice as often as it fails. Find the chance that in the next six trials, there shall be at least four successes.
(a) $\dfrac{233}{729}$
(b) $\dfrac{64}{729}$
(c) $\dfrac{496}{729}$
(d) $\dfrac{432}{729}$

Answer 1

Hint: The given question is a case of binomial probability distribution so we will use probability formula for a binomial distribution and that is $P\left( X=r \right)={}^{n}{{C}_{r}}{{\left( p \right)}^{r}}{{\left( q \right)}^{n-r}}\text{ , }r=0,1,2,3.....n$ , where $p$ is the probability of success and $q$ is the probability of failure and $n$ is the total number of trials and $r$ is the number of times we succeeded. After that, we will solve it with suitable values of $r$ as per the given data to get the correct answer.

Complete step-by-step answer:

Given:
There is an experiment which succeeds twice as often as it fails. This means that the probability of success of the experiment will be twice the probability of failure.
Now, let the probability of success of the experiment is $p$ and the probability of failure is $q$ . Then,
$\begin{align}
  & p+q=1 \\
 & p=2q \\
 & \text{Then, }2q+q=1 \\
 & \Rightarrow 3q=1 \\
 & \Rightarrow q=\dfrac{1}{3}\text{ and }p=\dfrac{2}{3} \\
\end{align}$
Now, by solving the above equations we got the probability of success of the experiment $p=\dfrac{2}{3}$ and probability of failure of the experiment $q=\dfrac{1}{3}$ . Before we proceed we should know the binomial distribution probability formula. It is written below:

Binomial Distribution: If a random variable $X$ which takes values $0,1,2,3............n$ follows a binomial distribution. Then, probability distribution function is given by,
$P\left( X=r \right)={}^{n}{{C}_{r}}{{\left( p \right)}^{r}}{{\left( q \right)}^{n-r}}\text{ , }r=0,1,2,3.....n$
Where, $p$ is the probability of success and $q$ is the probability of failure and $n$ is the total number of trials and $r$ is the number of times we succeeded.
Now, in our problem, we have a total of 6 trials and we have to find the probability of at least 4 successes. Which means, $n=6$ and $X=4,5,6$ . And as we have calculated above that $p=\dfrac{2}{3}$ and $q=\dfrac{1}{3}$ . For the probability of at least four successes, we have to add the probability for the values $X=4,5,6$ of the random variable. Then,
$\begin{align}
  & P\left( X=4 \right)+P\left( X=5 \right)+P\left( X=6 \right) \\
 & \Rightarrow {}^{6}{{C}_{4}}{{\left( p \right)}^{4}}{{\left( q \right)}^{6-4}}+{}^{6}{{C}_{5}}{{\left( p \right)}^{5}}{{\left( q \right)}^{6-5}}+{}^{6}{{C}_{6}}{{\left( p \right)}^{6}}{{\left( q \right)}^{6-6}} \\
 & \Rightarrow \dfrac{6\times 5}{2}\times {{\left( \dfrac{2}{3} \right)}^{4}}\times {{\left( \dfrac{1}{3} \right)}^{2}}+6\times {{\left( \dfrac{2}{3} \right)}^{5}}\times {{\left( \dfrac{1}{3} \right)}^{1}}+1\times {{\left( \dfrac{2}{3} \right)}^{6}}\times {{\left( \dfrac{1}{3} \right)}^{0}} \\
 & \Rightarrow \dfrac{240}{729}+\dfrac{192}{729}+\dfrac{64}{729} \\
 & \Rightarrow \dfrac{496}{729} \\
\end{align}$
Thus, as $P\left( X=4 \right)+P\left( X=5 \right)+P\left( X=6 \right)=\dfrac{496}{729}$ is the probability of at least four success in trials.
Hence, (c) is the correct option.

Note: Here, the student should apply the binomial distribution probability formula with proper values and should add the probabilities for the 4, 5 and 6 successes because in the question we have to find the probability for at least four successes and not only 4 successes.