Question

A coin is biased so that the head is three times more likely to occur than the tail. If the coin is tossed twice, find the probability distribution of the number of tails.
[a] $\dfrac{9}{16},\dfrac{3}{16},\dfrac{1}{16}$
[b] $\dfrac{8}{16},\dfrac{3}{16},\dfrac{1}{16}$
[c] $\dfrac{9}{16},\dfrac{3}{8},\dfrac{1}{16}$
[d] $\dfrac{9}{16},\dfrac{3}{16},\dfrac{1}{8}$

Answer 1

Hint: Use the fact that if A and B are independent events then $P\left( A\bigcap B \right)=P\left( A \right)P\left( B \right)$. Assume A be the event of getting a tail on the first try and B the event of getting a tail on the second try. Use the fact that A, B are independent since the trials are independent hence find the probability of each of the events $A\bigcap B,A\bigcap B',A'\bigcap B,A'\bigcap B'$. Hence determine the probability of getting no tail, one tail or two tails. Hence determine the probability distribution of the number of tails.

Complete step-by-step answer:
Let A be the event of getting a tail on the first try and B the event of getting a tail on the second try. Let X be the random variable: The number of tails obtained.
Clearly, X = 0 is the event $A'\bigcap B'$, X =1 is the event $\left( A\bigcap B' \right)\bigcup \left( A'\bigcap B \right)$ and X=2 is the event $A\bigcap B$.
Now, we have A and B are independent events, since the trials are independent.
Since A and B are independent events, then so are the events A’ and B, A and B’, and A’ and B’.
Hence, we have
$P\left( A\bigcap B \right)=P\left( A \right)P\left( B \right)$
Now, we have P(A) is the probability of getting a tail on the first try. Since the head is 3 times as likely to occur a tail. Hence, we have
$P\left( A \right)=P\left( B \right)=\dfrac{1}{3+1}=\dfrac{1}{4}$
Hence, we have $P\left( A' \right)=1-\dfrac{1}{4}=\dfrac{3}{4}=P\left( B' \right)$
Hence, we have
$\begin{align}
  & P\left( A\bigcap B \right)=\dfrac{1}{4}\times \dfrac{1}{4}=\dfrac{1}{16} \\
 & P\left( A'\bigcap B \right)=\dfrac{3}{4}\times \dfrac{1}{4}=\dfrac{3}{16} \\
 & P\left( A\bigcap B' \right)=\dfrac{1}{4}\times \dfrac{3}{4}=\dfrac{3}{16} \\
 & P\left( A'\bigcap B' \right)=\dfrac{3}{4}\times \dfrac{3}{4}=\dfrac{9}{16} \\
\end{align}$
We know that if A and B are mutually exclusive events, then $P\left( A\bigcup B \right)=P\left( A \right)+P\left( B \right)$
Now, we have
$\begin{align}
  & P\left( X=0 \right)=P\left( A\bigcup B \right)=\dfrac{9}{16} \\
 & P\left( X=1 \right)=P\left( \left( A'\bigcap B \right)\bigcup \left( A\bigcap B' \right) \right) \\
\end{align}$
Since $A'\bigcap B$ and $A\bigcap B'$ are independent events, we have
$\begin{align}
  & P\left( X=1 \right)=P\left( A'\bigcap B \right)+P\left( A\bigcap B' \right)=\dfrac{3}{16}+\dfrac{3}{16}=\dfrac{6}{16} \\
 & P\left( X=2 \right)=P\left( A'\bigcap B' \right)=\dfrac{1}{16} \\
\end{align}$
Hence the probability of distribution of the number of tails is as shown below
Hence option [c] is correct.

Note: Verification:
We can verify our answer is correct by checking that if the sum of all the probabilities in the probability distribution is 1.
$\dfrac{9}{16}+\dfrac{6}{16}+\dfrac{1}{16}=1$
Hence our answer is verified to be correct.
Also, you can observe none of the options except option [c] satisfies the above property.
Hence option [c] is correct.