Question

A biased coin is tossed twice. The probability of head is twice the tail. The PDF of the number of heads is

$ x $	0	1	2
$ P\left( x \right) $	$ \dfrac{a}{d} $	$ \dfrac{b}{d} $	$ \dfrac{c}{d} $

Then the values of a, b, c, d are
A. a = 1, b = 2, c = 3, d = 4
B. a = 1, b = 4, c = 4, d = 9
C. a = 1, b = 4, c = 4, d = 10
D. a = 1, b = 2, c = 1, d = 4

Answer 1

Hint: To solve the given question, we need to know what a binomial distribution is. A binomial distribution can be thought of as simply the probability of a success or a failure outcome in an experiment. If we have n as the number of trials, x as the number of successes desired, p as the probability of getting a success in one trial and q as the probability of getting a failure in one trial, which is always 1 -p, then the probability distribution, $ P\left( x \right)=n{{C}_{x}}{{p}^{x}}{{q}^{n-x}} $ . We will use this concept to get our desired answer.

Complete step-by-step answer:
We have been given the question that a biased coin is tossed twice and the probability of head is twice the tail. We have been given a PDF of the number of heads and have been asked to find the values of a, b, c, d. Here, we are given the experiment as tossing a biased coin twice, so the number of trials, n = 2 times. Now, let us consider the event of getting head as a success. Therefore, we get,
The probability of getting head in a trial = $ p $ , and the probability of getting tail in a trial = $ q $ .
We are also given that the probability of getting a head is twice that of the tail, so it means, $ p=2q $ or $ q=\dfrac{p}{2} $ . But we know that the total probability should be equal to 1. We have a binomial distribution here with two events, getting a head or a tail. So, the total probability will be,
 $ p+q=1 $
We know that $ p=2q $ , so we will substitute it in the above equation and get,
 $ 2q+q=1\Rightarrow q=\dfrac{1}{3} $
So, it means, we can write the value of $ p $ as $ p=2q=\dfrac{2}{3} $ . Now we will find the required values as per given in the question one by one.
(i) When we have the value of $ x=0 $
So, the probability distribution can be found out by using the formula, $ P\left( x \right)=n{{C}_{x}}{{p}^{x}}{{q}^{n-x}} $ . Here, we have $ x=0,n=2,p=\dfrac{2}{3},q=\dfrac{1}{3} $ . Substituting these values in the formula, we get,
 $ \begin{align}
  & P\left( 0 \right)=2{{C}_{0}}{{\left( \dfrac{2}{3} \right)}^{0}}{{\left( \dfrac{1}{3} \right)}^{2-0}} \\
 & \Rightarrow P\left( 0 \right)=2{{C}_{0}}{{\left( \dfrac{2}{3} \right)}^{0}}{{\left( \dfrac{1}{3} \right)}^{2}} \\
 & \Rightarrow P\left( 0 \right)=\dfrac{2!}{2!0!}\times \left( 1 \right)\times \left( \dfrac{1}{9} \right) \\
 & \Rightarrow P\left( 0 \right)=1\times 1\times \dfrac{1}{9} \\
 & \Rightarrow P\left( 0 \right)=\dfrac{1}{9} \\
\end{align} $
(ii) When we have the value of $ x=1 $
So, here we will have the values as, $ x=1,n=2,p=\dfrac{2}{3},q=\dfrac{1}{3} $ . So, applying these in the formula, we will get,
 $ \begin{align}
  & P\left( 1 \right)=2{{C}_{1}}{{\left( \dfrac{2}{3} \right)}^{1}}{{\left( \dfrac{1}{3} \right)}^{2-1}} \\
 & \Rightarrow P\left( 1 \right)=2{{C}_{1}}{{\left( \dfrac{2}{3} \right)}^{1}}{{\left( \dfrac{1}{3} \right)}^{1}} \\
 & \Rightarrow P\left( 1 \right)=\dfrac{2!}{1!1!}\times \dfrac{2}{3}\times \dfrac{1}{3} \\
 & \Rightarrow P\left( 1 \right)=\dfrac{2\times 2}{9} \\
 & \Rightarrow P\left( 1 \right)=\dfrac{4}{9} \\
\end{align} $
(iii) When we have the value of $ x=2 $
Now, we have the values as, $ x=2,n=2,p=\dfrac{2}{3},q=\dfrac{1}{3} $ , so we get,
 $ \begin{align}
  & P\left( 2 \right)=2{{C}_{2}}{{\left( \dfrac{2}{3} \right)}^{2}}{{\left( \dfrac{1}{3} \right)}^{2-2}} \\
 & \Rightarrow P\left( 2 \right)=2{{C}_{2}}{{\left( \dfrac{2}{3} \right)}^{2}}{{\left( \dfrac{1}{3} \right)}^{0}} \\
 & \Rightarrow P\left( 2 \right)=\dfrac{2!}{2!0!}\times \dfrac{4}{9}\times 1 \\
 & \Rightarrow P\left( 2 \right)=\dfrac{4}{9} \\
 & \Rightarrow P\left( 1 \right)=\dfrac{4}{9} \\
\end{align} $
Thus, we get the probability distribution of the number of heads as,

$ x $	0	1	2
$ P\left( x \right) $	$ \dfrac{1}{9} $	$ \dfrac{4}{9} $	$ \dfrac{4}{9} $

So, we get the values of a = 1, b = 4, c = 4, and d = 9
So, the correct answer is “Option B”.

Note: We must note that if the coin was unbiased, then we would have taken the probability of getting head, $ p=\dfrac{1}{2} $ and the probability of getting a tail, $ q=\dfrac{1}{2} $ and the total probability would be equal to 1. We should also remember that binomial distribution can be considered only if there are two events, one is a success and the other is a failure as it was the case with the given question.