Question

A bag contains 7 red, 5 white and 8 black balls. If four balls are drawn one by one with replacement, the probability that any two are white is $\dfrac{{{3}^{a}}}{{{2}^{b}}}$ then value of a+b is ....

Answer 1

Hint: In this question, we are given a number of red, white and black balls. We have to find the probability of two balls drawn to be white when four balls are drawn one by one with replacement. For this, we will use binomial distribution to find probability. Binomial distribution for occurring of x event is given as:
\[P\left( x \right)={}^{n}{{C}_{x}}{{\left( p \right)}^{x}}{{\left( q \right)}^{n-x}}\]
Where n is the number of trials (or number being sampled), x is the number of successes in one trial and q is the probability of getting failure in one trial. Here, q can be calculated as q = 1-p.

Complete step by step answer:
Here, we are given a number of red balls as 7, number of white balls as 5 and number of black balls as 8. So the total number of balls will be 7+5+8 = 20.
If one ball is picked from these balls then probability of getting white ball will be equal to
\[\text{Probability}=\dfrac{\text{Number of white balls}}{\text{Total number of balls}}=\dfrac{5}{20}=\dfrac{1}{4}\]
Now, let X event be that white ball is chosen. Since $\dfrac{1}{4}$ is probability of success for event X, therefore $p=\dfrac{1}{4}$.
Since q is probability of failure for event X, so q is given by 1-p, therefore $q=1-\dfrac{1}{4}=\dfrac{3}{4}$.
So $q=\dfrac{3}{4}$.
Now total chances taken are 4, therefore n = 4. We want two white balls, therefore x = 2.
As we know, binomial distribution is given by
\[P\left( X=x \right)={}^{n}{{C}_{x}}{{\left( p \right)}^{x}}{{\left( q \right)}^{n-x}}\]
Where n is the number of trials (or number being sampled), x is the number of desired success, p is probability of getting success in one trial and q is the probability of getting failure in one trial. Here, we have found, n = 4, x = 2, $p=\dfrac{1}{4}\text{ and }q=\dfrac{3}{4}$. So, P(X) becomes
\[\begin{align}
  & P\left( X=2 \right)={}^{4}{{C}_{2}}{{\left( \dfrac{1}{4} \right)}^{2}}{{\left( \dfrac{3}{4} \right)}^{4-2}} \\
 & \Rightarrow \dfrac{4!}{2!2!}\times \dfrac{1}{4}\times \dfrac{1}{4}\times {{\left( \dfrac{3}{4} \right)}^{2}} \\
 & \Rightarrow \dfrac{4\times 3\times 2\times 3\times 3}{2\times 2\times 4\times 4\times 4\times 4} \\
 & \Rightarrow \dfrac{27}{128} \\
 & \Rightarrow \dfrac{{{3}^{3}}}{{{2}^{7}}} \\
\end{align}\]
We are given probability as $\dfrac{{{3}^{a}}}{{{2}^{b}}}$ therefore, comparing it with $\dfrac{{{3}^{3}}}{{{2}^{7}}}$ we get a = 3 and b = 7.
Adding a and b we get a+b = 3+7 = 10.

So, the correct answer is “10”.

Note: Students should carefully apply the formula of binomial distribution. Students can get confused between n and x, n is the total ways of selecting and x is the desired ways of selecting. Determine value of p and q carefully using proper data. While calculating ${}^{n}{{C}_{r}}$ use the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.