Question

How do you find the probability of at least two successes when $n$ independent Bernoulli trials are carried out with probability of success $p$?

Answer 1

Hint: In the problem they have mentioned that the probability distribution is the Bernoulli distribution. We know that the probability of an event in Bernoulli distribution is given by $P\left( X=x \right){{=}^{n}}{{C}_{x}}~{{p}^{x}}~{{q}^{n-x}}$. Where $n$ is the repeated times and $p$ is the probability of success and $q$ is the probability of the failure and it will be equal to one less than the probability of success mathematically $q=1-p$. In the problem they have asked to calculate the probability of at least two successes so we will calculate the probabilities of the required events using the above equation. Now we will calculate the required probability from the calculated probabilities of the events.

Complete step by step answer:
Given distribution is Bernoulli Distribution.
Number of independent trains is $n$.
Probability of success is $p$.
Now the probability of failure is equal to the one less than the probability of success. I.e.,
$q=1-p$.
Now the probability of an event $x$ in Bernoulli distribution can be calculated by
$\begin{align}
  & P\left( X=x \right)={}^{n}{{C}_{x}}{{p}^{x}}{{q}^{n-x}} \\
 & \Rightarrow P\left( X=x \right)={}^{n}{{C}_{x}}{{p}^{x}}{{\left( 1-p \right)}^{n-x}} \\
\end{align}$
Now we need to calculate the probability of at least two successes mathematically we can write it as $P\left( X\ge 2 \right)$. In the problem we don’t have the number of trials done, so we don’t have how many events are in the sample space. So, it is not possible to calculate the value of $P\left( X\ge 2 \right)$ directly for this we will calculate the value of $P\left( X\ge 2 \right)$ indirectly from the below equation
$P\left( X\ge 2 \right)=1-P\left( X=0 \right)-P\left( X=1 \right)$
Using the formula $P\left( X=x \right)={}^{n}{{C}_{x}}{{p}^{x}}{{\left( 1-p \right)}^{n-x}}$ in the above equation, then we will get
$\Rightarrow P\left( X\ge 2 \right)=1-{}^{n}{{C}_{0}}{{p}^{0}}{{\left( 1-p \right)}^{n-0}}-{}^{n}{{C}_{1}}{{p}^{1}}{{\left( 1-p \right)}^{n-1}}$
We know that ${}^{n}{{C}_{0}}=1$, ${}^{n}{{C}_{1}}=n$, ${{p}^{0}}=1$, then we will get
$\Rightarrow P\left( X\ge 2 \right)=1-{{\left( 1-p \right)}^{n}}-np{{\left( 1-p \right)}^{n-1}}$
Taking common ${{\left( 1-p \right)}^{n-1}}$ in the above equation, then we will get
$\begin{align}
  & \Rightarrow P\left( X\ge 2 \right)=1-{{\left( 1-p \right)}^{n-1}}\left[ \left( 1-p \right)-np \right] \\
 & \Rightarrow P\left( X\ge 2 \right)=1-{{\left( 1-p \right)}^{n-1}}\left[ 1-p-np \right] \\
 & \Rightarrow P\left( X\ge 2 \right)=1-{{\left( 1-p \right)}^{n-1}}\left[ 1-p\left( n+1 \right) \right] \\
\end{align}$

Hence the required probability is $P\left( X\ge 2 \right)=1-{{\left( 1-p \right)}^{n-1}}\left[ 1-p\left( n+1 \right) \right]$.

Note: We have different distributions for the probabilities. For the Bernoulli distribution we have used this formula but when they have mentioned different distributions, then we need to use the different formula.