Question

In a throw of dice the probability of getting one in even number of throws is

Answer 1

Hint: In this question, we need to find the probability of getting one in even number of throws. For that, we first find the probability of getting one in the even number of throws that is second throw, fourth throw, the sixth throw and so on then find the required probability.

Formula used:
We have been using the following formulas:
1.To find the sum of infinite series \[{S_n} = \dfrac{a}{{1 - r}}\] where \[r = \dfrac{b}{a}\]

Complete step-by-step solution:
Now we know that the die has a total of six numbers. So, the total number of outcomes \[ = 6\]
Thus, the set of numbers \[ = 1,2,3,4,5,\,\,and\,\,6\]
Now we know that even numbers are those which are divisible by \[2\]
Thus, we can say 2, 4, 6, 8 and so are even numbers.
Now the probability of getting one in the second throw is:
\[\left( {\dfrac{5}{6}} \right)\left( {\dfrac{1}{6}} \right)\]
Now the probability of getting one in the fourth throw is:
\[{\left( {\dfrac{5}{6}} \right)^3}\left( {\dfrac{1}{6}} \right)\]
Now the probability of getting one in the sixth throw is:
\[{\left( {\dfrac{5}{6}} \right)^5}\left( {\dfrac{1}{6}} \right)\]
And so on
Hence, the required probability is \[ = P\left( 2 \right) + P\left( 4 \right) + P\left( 6 \right)...\]
\[
  P\left( 2 \right) + P\left( 4 \right) + P\left( 6 \right)... = {\left( {\dfrac{5}{6}} \right)^1}\left( {\dfrac{1}{6}} \right) + {\left( {\dfrac{5}{6}} \right)^3}\left( {\dfrac{1}{6}} \right) + {\left( {\dfrac{5}{6}} \right)^5}\left( {\dfrac{1}{6}} \right)...\infty \\
   = \left( {\dfrac{1}{6}} \right)\left[ {{{\left( {\dfrac{5}{6}} \right)}^1} + {{\left( {\dfrac{5}{6}} \right)}^3} + {{\left( {\dfrac{5}{6}} \right)}^5}} \right]...\infty
 \]
Now, by using the sum of infinite geometric series \[{S_n} = \dfrac{a}{{1 - r}}\] where \[ r = \dfrac{b}{a}\], the required probability is
\[
  {S_n} = \left( {\dfrac{1}{6}} \right)\left[ {\dfrac{{\dfrac{5}{6}}}{{1 - \dfrac{{25}}{{36}}}}} \right] \\
   = \left( {\dfrac{1}{6}} \right)\left[ {\dfrac{{\dfrac{5}{6}}}{{\dfrac{{36 - 25}}{{36}}}}} \right] \\
   = \left( {\dfrac{1}{6}} \right)\left[ {\dfrac{{\dfrac{5}{6}}}{{\dfrac{{11}}{{36}}}}} \right]
 \]
Now, by simplifying the above expression, we get the required probability:

\[
  {S_n} = \left( {\dfrac{1}{6}} \right)\left[ {\dfrac{5}{6} \times \dfrac{{36}}{{11}}} \right] \\
   = \left( {\dfrac{1}{6}} \right)\left[ {\dfrac{{30}}{{11}}} \right] \\
   = \dfrac{5}{{11}}
 \]
Therefore, the required probability is \[\dfrac{5}{{11}}\]

Note: Natural numbers are divided into two groups: even numbers and odd numbers. Even numbers are those which are completely divisible by 2, whereas odd numbers are those that are not eventually completely divisible by 2.