Question

A biased coin with probability P, (0 < p < 1) of heads is tossed until a head appear for the first time. If the probability that the number of tosses required is even is $\dfrac{2}{5}$ then P =
(a) $\dfrac{2}{5}$
(b) $\dfrac{3}{5}$
(c) $\dfrac{2}{3}$
(d) $\dfrac{1}{3}$

Answer 1

Hint: In this problem, the probability of the number of tosses of coins is given. By using this we try to make a series of chances of all even toss i.e. chance of doing 2 tosses, chance of doing 4 tosses, chance of doing 6 tosses, and so on. Using this we can easily find the value of P.

Complete step-by-step answer:
One of the important aspects of mathematical operations is to predict the possibility of occurrence of an event. This introduces the concept of probability in mathematical models.
The probability of success is p and failure is 1 - p. Now, considering the problem statement, the appearance of head in even time is
The possibility of 2 tosses i.e. failure in first and success at next. It can be mathematically shown as multiplication of failure (1 – p) with success p $=\left( 1-p \right)\times p$
The possibility of 4 tosses i.e. failure in first, second, third and success at next $=\left( 1-p \right)\times \left( 1-p \right)\times \left( 1-p \right)\times p$
The possibility of 6 tosses i.e. failure in first, second, third, fourth, fifth and success at next $=\left( 1-p \right)\times \left( 1-p \right)\times \left( 1-p \right)\times \left( 1-p \right)\times \left( 1-p \right)\times p$
Now, the probability that the number of coins is even = The possibility of 2 tosses + The possibility of 4 tosses + The possibility of 6 tosses……. to infinity.
$P\left( E \right)=\left( 1-p \right)\times p+\left( 1-p \right)\times \left( 1-p \right)\times \left( 1-p \right)\times p+\left( 1-p \right)\times \left( 1-p \right)\times \left( 1-p \right)\times \left( 1-p \right)\times \left( 1-p \right)\times p...\infty $
One useful formula for solving the following expression is sum of infinite series which could be stated as:
$a\left( 1+r+{{r}^{2}}+{{r}^{3}}............\infty \right)=\dfrac{a}{1-r}\ldots \left( 1 \right)$.
Taking, $\left( 1-p \right)\times p$ common from P(E), we get
\[\begin{align}
  & =\left( 1-p \right)\times p\left[ 1+\left( 1-p \right)\times \left( 1-p \right)+\left( 1-p \right)\times \left( 1-p \right)\times \left( 1-p \right)\times \left( 1-p \right)+......\infty \right] \\
 & =\left( 1-p \right)\times p\left[ 1+{{\left( 1-p \right)}^{2}}+{{\left( 1-p \right)}^{4}}+......\infty \right]\ldots \left( 2 \right) \\
\end{align}\]
On comparing equation (1) and equation (2), \[r={{\left( 1-p \right)}^{2}}\text{ and }a=\left( 1-p \right)\times p\], so on replacing in final simplified expression, we get
$\dfrac{a}{1-r}=\left( \dfrac{\left( 1-p \right)\times p}{1-{{\left( 1-p \right)}^{2}}} \right)$
Now, as given in the problem, the probability is $\dfrac{2}{5}$. Therefore,
$\Rightarrow \left( \dfrac{\left( 1-p \right)\times p}{{{1}^{2}}-{{\left( 1-p \right)}^{2}}} \right)=\dfrac{2}{5}$
Now, by using the identity of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$, we get
$\begin{align}
  & \Rightarrow \left( \dfrac{\left( 1-p \right)\times p}{\left( 1-1+p \right)\left( 1+1-p \right)} \right)=\dfrac{2}{5} \\
 & \Rightarrow \left( \dfrac{\left( 1-p \right)\times p}{\left( p \right)\left( 2-p \right)} \right) \\
 & \Rightarrow \left( \dfrac{1-p}{2-p} \right)=\dfrac{2}{5} \\
 & \Rightarrow 5-5p=4-2p \\
 & \Rightarrow 1=3p \\
 & \Rightarrow p=\dfrac{1}{3} \\
\end{align}$
On, solving the above equation, we get the value of p:
P$=\dfrac{1}{3}$
Hence, If the probability that the number of tosses required is even is $\dfrac{2}{5}$ then the P$=\dfrac{1}{3}$.
Therefore, option (d) is correct.


Note: The key step for solving this problem is the knowledge of multiple probabilities of an event. If a coin is tossed multiple times, then events can be expressed in terms of a series. By identifying the type of series, we can easily evaluate the value of the required quantity. Hence knowledge of sequence and series is also important.