Question

A biased coin with the probability \[p,{\text{ }}0 < p < 1\] of heads is tossed until a head appears for the first time. If the probability that the number of tosses required is even is \[\dfrac{2}{5}\]. Find the value of \[p\].

Answer 1

Hint: Probability means possibility. Here, we use the concept of probability. When an unbiased coin is tossed, there are only two possibilities of outcomes i.e., HEAD and TAIL.
Here, in the question a coin is tossed until it results in getting a HEAD, for which we will use the basic formulae of the probability as:
\[
  {\text{Probability of an event}} = \dfrac{{Favourable{\text{ }}cases}}{{Total{\text{ }}cases}} \\
  P(event{\text{ }}happening) + P(event{\text{ }}not{\text{ }}happening) = 1 \\
\]
First we have to evaluate the probability of getting a head in two simultaneous tosses, then for four simultaneous tosses and then for six tosses and then we could eventually derive an expression whose result will find the required probability.


Complete step by step solution:
A biased coin with probability \[p,{\text{ }}0 < p < 1\] is tossed until a head appears for the first time.
$P(Head) = p$
Then,
$
  P(Tail) = 1 - P(Head) \\
   = 1 - p \\
 $
The outcome of being a head in two tosses implies that the first toss results in the tail, and the second one is head. Then, ${P_2} = \left( {1 - p} \right) \times p$
The outcome of being a head in four tosses implies that the first three tosses are tail and the fourth toss is head. Then, \[{P_4} = \left( {1 - p} \right) \times \left( {1 - p} \right) \times \left( {1 - p} \right) \times p\]
The outcome of being a head in six tosses implies that the first five tosses are tail and the sixth toss is head. Then, \[{P_6} = \left( {1 - p} \right) \times \left( {1 - p} \right) \times \left( {1 - p} \right) \times \left( {1 - p} \right) \times \left( {1 - p} \right) \times p\]
The probability that the number of tosses required is even.
According to the question, the probability that the number of tosses required is even is \[\dfrac{2}{5}\] which is followed as:
$
  P = {P_2} + {P_4} + {P_6} + ... \\
  \dfrac{2}{5} = {(1 - p)^1} \times p + {(1 - p)^3} \times p + {(1 - p)^5} \times p + ....) \\
  \dfrac{2}{5} = {(1 - p)^1} \times p\left[ {1 + {{(1 - p)}^2} + {{(1 - p)}^4} + ....} \right] - - - - (i) \\
 $
\[1 + {(1 - p)^2} + {(1 - p)^4} + ......\] is an infinite geometric progression.

The Sum of terms in the infinite geometric progression is given as: \[\dfrac{a}{{1 - r}},here{\text{ }}a = 1{\text{ }}and{\text{ }}r = {(1 - p)^2}\]
\[S = \dfrac{1}{{1 - {{(1 - p)}^2}}}\]
Substituting in equation (i) as:
\[
  {(1 - p)^1} \times p\left[ {\dfrac{1}{{1 - {{(1 - p)}^2}}}} \right] = \dfrac{2}{5} \\
  \dfrac{{p - {p^2}}}{{1 - 1 - {p^2} + 2p}} = \dfrac{2}{5} \\
  5p - 5{p^2} = - 2{p^2} + 4p \\
  p = 3{p^2} \\
  1 = 3p \\
  p = \dfrac{1}{3} \\
 \]

Therefore, the value of $p$ $ = \dfrac{1}{3}$


Note: In such types of questions, which involves the concept of probability, knowledge about the definition is needed. Sometimes probability questions may be combined with the concept of infinite geometric progression. Follow the conditions in the question and solve accordingly to get the required value.