Question

If the probability of rain on any given day in Pune city is \[50\%\]. What is the probability that it rains on exactly 3 days in a 5 day period?

Answer 1

Hint: We have given the probability of rain on any given day in Pune city is \[50\%\]. We calculate the total number of possible outcomes. Then, we calculate the number of ways to choose 3 days in a 5 day period. Then, we calculate the probability by using the general formula of probability, which is given by
$P\left( A \right)=\dfrac{n\left( E \right)}{n\left( S \right)}$
Where, A is an event,
\[n\left( E \right)=\] Number of favorable outcomes and $n\left( S \right)=$ number of total possible outcomes

Complete step-by-step answer:
We have given that the probability of rain on any given day in Pune city is \[50\%\].
We have to find the probability that it rains on exactly 3 days in a 5 day period.
Now, as given in the question probability of rain on any given day in Pune city is \[50\%\]. It means the probability of no rain is also \[50\%\].
We can write it as Probability of rain on a day is $\dfrac{1}{2}$.
Also, probability of no rain is $\dfrac{1}{2}$.
Now, the number of ways to choose exactly 3 days in a 5 day period will be
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
Where, $n=$ number of items/objects
And $r=$ number of items/objects being chosen at a time
So, number of ways to choose exactly 3 days in a 5 day period will be
\[\begin{align}
  & {}^{5}{{C}_{3}}=\dfrac{5!}{3!\left( 5-3 \right)!} \\
 & {}^{5}{{C}_{3}}=\dfrac{5\times 4\times 3!}{3!\left( 2 \right)!} \\
 & {}^{5}{{C}_{3}}=\dfrac{5\times 4}{2\times 1} \\
 & {}^{5}{{C}_{3}}=\dfrac{20}{2} \\
 & {}^{5}{{C}_{3}}=10 \\
\end{align}\]
Now, the probability that it rains on exactly 3 days in a 5 day period will be
\[\begin{align}
  & \Rightarrow {}^{5}{{C}_{3}}\times {{\left( \dfrac{1}{2} \right)}^{3}}\times {{\left( \dfrac{1}{2} \right)}^{2}} \\
 & \Rightarrow 10\times \dfrac{1}{8}\times \dfrac{1}{4} \\
 & \Rightarrow \dfrac{10}{32} \\
 & =\dfrac{5}{16} \\
\end{align}\]
So, the probability that it rains on exactly 3 days in a 5 day period is $\dfrac{5}{16}$.

Note: Alternatively we can solve this question by following way-
Total number of days we have is $5$.
So, the total number of possible outcomes will be ${{2}^{5}}=32$
Number of ways to choose 3 days in a period of 5 day will be $=10$ (As calculated above)
Now, the probability that it rains on exactly 3 days in a 5 day period will be
$P\left( A \right)=\dfrac{n\left( E \right)}{n\left( S \right)}$
$\begin{align}
  & P\left( A \right)=\dfrac{10}{32} \\
 & P\left( A \right)=\dfrac{5}{16} \\
\end{align}$