Question

In a throw of a die the probability of getting one in an even number of throws is $\dfrac{\lambda }{{11}}$, then find $\lambda $.

Answer 1

Hint:Find the probability of getting 1 and the probability of not getting 1. Now find the probability of getting 1 for the first time in the second throw, then find the probability of getting 1 for the first time in the fourth throw, and so on. Add these probabilities to get the answer.

Formula Used: Sum to infinity of a geometric progression is $\dfrac{a}{{1 - r}}$ where $a,r$ are the first term and the constant ratio respectively and $r < 1$.

Complete step by step Solution:
Sample space = $\left\{ {\left( 1 \right),\left( 2 \right),\left( 3 \right),\left( 4 \right),\left( 5 \right),\left( 6 \right)} \right\}$
Therefore, the total number of outcomes is 6
Number of favorable outcomes = 1
Let the event of getting one by event A.
$P\left( A \right) = \dfrac{1}{6}$
$P\left( {\overline A } \right) = 1 - \dfrac{1}{6} = \dfrac{5}{6}$
Let the event of getting one in an even number of throws be event X.
To find the probability of getting one in an even number of throws we need to add the probability of getting one for the first time in the second throw, probability of getting one for the first time in the fourth throw, probability of getting one for the first time in the sixth throw and so on.
Let the event of getting one for the first time in the second throw be event B.
$P\left( B \right) = P\left( {\overline A } \right)P\left( A \right) = \dfrac{5}{6}.\dfrac{1}{6}$
Let the event of getting one for the first time in the fourth throw be event C.
$P\left( C \right) = P\left( {\overline A } \right)P\left( {\overline A } \right)P\left( {\overline A } \right)P\left( A \right) = \dfrac{{{5^3}}}{{{6^3}}}.\dfrac{1}{6}$
Let the event of getting one for the first time in the sixth throw be event D.
$P\left( D \right) = P\left( {\overline A } \right)P\left( {\overline A } \right)P\left( {\overline A } \right)P\left( {\overline A } \right)P\left( {\overline A } \right)P\left( A \right) = \dfrac{{{5^5}}}{{{6^5}}}.\dfrac{1}{6}$
$P\left( X \right) = P\left( B \right) + P\left( C \right) + P\left( D \right) + .......$
$P\left( X \right) = \dfrac{5}{6}.\dfrac{1}{6} + \dfrac{{{5^3}}}{{{6^3}}}.\dfrac{1}{6} + \dfrac{{{5^5}}}{{{6^5}}}.\dfrac{1}{6} + .......$
Solving further, we get
$P\left( X \right) = \dfrac{5}{6}.\dfrac{1}{6}\left[ {\dfrac{{{5^0}}}{{{6^0}}} + \dfrac{{{5^2}}}{{{6^2}}} + \dfrac{{{5^4}}}{{{6^4}}} + .......} \right]$
$P\left( X \right) = \dfrac{5}{6}.\dfrac{1}{6}\left[ {\dfrac{1}{{1 - \dfrac{{{5^2}}}{{{6^2}}}}}} \right]$
$P\left( X \right) = \dfrac{5}{6}.\dfrac{1}{6}\left[ {\dfrac{{36}}{{11}}} \right] = \dfrac{5}{{36}}.\dfrac{{36}}{{11}}$
$P\left( X \right) = \dfrac{5}{{11}}$
Therefore, the value of $\lambda $ is 5.

Note: An alternate method of simplifying from $P\left( X \right) = \dfrac{5}{6}.\dfrac{1}{6} + \dfrac{{{5^3}}}{{{6^3}}}.\dfrac{1}{6} + \dfrac{{{5^5}}}{{{6^5}}}.\dfrac{1}{6} + .......$:
$P\left( X \right) = \dfrac{5}{{{6^2}}} + \dfrac{{{5^3}}}{{{6^4}}} + \dfrac{{{5^5}}}{{{6^6}}} + .....$.
We can see that this is a case of the sum to infinity of a geometric progression where the first term is $\dfrac{5}{{{6^2}}}$ and the common difference is $\dfrac{{{5^2}}}{{{6^2}}}$.
$P\left( X \right) = \left[ {\dfrac{{\dfrac{5}{{{6^2}}}}}{{1 - \dfrac{{{5^2}}}{{{6^2}}}}}} \right] = \dfrac{5}{{11}}$