Question

A biased coin with probability P, 0 < P < 1, of heads is tossed until a head appears for the first time. If the probability that the number of tosses required is even is 2/5, then P =.......
$\left( A \right)$ 2/5
$\left( B \right)$ 2/3
$\left( C \right)$ 1/3
$\left( D \right)$ 3/5

Answer 1

Hint – In this particular type of question use the concept that when a biased coin is tossed the probability of getting the head is given as $P\left( {x = n} \right) = P{\left( {1 - P} \right)^{n - 1}}$, where n = positive integers starting from 1 so use this concept to reach the solution of the question.

Complete step-by-step answer:
Given data:
Probability that the number of tosses required is even is 2/5
$\Rightarrow P\left( E \right) = \dfrac{2}{5}$................... (1), Where E is the event that denotes the number of tosses required are even.
Let the number of tosses required = x.
So the probability to getting a head for the very first time when the biased coin is tossed is given as
$P\left( {x = n} \right) = P{\left( {1 - P} \right)^{n - 1}}$......................... (2), where n = positive integers starting from 1.
Now, let (E) be the event that denotes the number of tosses required are even.
So the set of even numbers = {2, 4, 6, 8,..........}
So, P(E) = \left\\{ {\left( {x = 2} \right) \cup \left( {x = 4} \right) \cup \left( {x = 6} \right) \cup ...............} \right\\}, where $\cup$ is the symbol of union.
$\Rightarrow P\left( E \right) = P\left( {x = 2} \right) + P\left( {x = 4} \right) + P\left( {x = 6} \right) + .............. + P\left( {x = \infty } \right)$, as there are infinite even terms possible.
Now from equation (2) we have,
$\Rightarrow P\left( E \right) = P{\left( {1 - P} \right)^{2 - 1}} + P{\left( {1 - P} \right)^{4 - 1}} + P{\left( {1 - P} \right)^{6 - 1}} + .............. + \infty$
$\Rightarrow P\left( E \right) = P{\left( {1 - P} \right)^1} + P{\left( {1 - P} \right)^3} + P{\left( {1 - P} \right)^5} + .............. + \infty$
Now as we see the above sum is the sum of the infinite terms G.P, whose first term (a) is P (1 – P), common ratio (r) is ${\left( {P - 1} \right)^2}$, and the number of terms is infinite.
Now the sum of the infinite terms G.P is given as, ${S_\infty } = \dfrac{a}{{1 - r}}$
So the sum is
$\Rightarrow {S_\infty } = P\left( E \right) = \dfrac{{P\left( {1 - P} \right)}}{{1 - {{\left( {1 - P} \right)}^2}}}$
Now simplify this we have,
$\Rightarrow P\left( E \right) = \dfrac{{P\left( {1 - P} \right)}}{{1 - {{\left( {1 - P} \right)}^2}}} = \dfrac{{P\left( {1 - P} \right)}}{{1 - 1 - {P^2} + 2P}} = \dfrac{{P\left( {1 - P} \right)}}{{P\left( {2 - P} \right)}} = \dfrac{{1 - P}}{{2 - P}}$
Now from equation (1) we have,
$\Rightarrow \dfrac{2}{5} = \dfrac{{1 - P}}{{2 - P}}$
Now solve for P we have,
$\Rightarrow 4 - 2P = 5 - 5P$
$\Rightarrow 5P - 2P = 5 - 4 = 1$
$\Rightarrow 3P = 1$
$\Rightarrow P = \dfrac{1}{3}$
So this is the required probability
Hence option (C) is the correct answer.

Note – Whenever we face such types of question the key concept we have to remember is that the probability that the number of tosses required is even is the sum of all the probabilities having even number which has infinite numbers as above and always recall the formula for sum of infinite terms of G.P then solve accordingly we will get the required answer.