Question

A fair dice is rolled till the number 1 appears on the top face of it. Find the probability that the dice is thrown an even number of times.

Answer 1

Hint: In the given question, we are required to find the probability of the event in which fair die is rolled till number $ 1 $ appears on top face of it and the number of throws of die is even. This problem is a typical example of a die problem involving the concepts of permutations and combinations and probability.

Complete step by step solution:
The probability of getting number $ 1 $ in single throw of a dice $ = \dfrac{1}{6} $
The probability of getting any number other than $ 1 $ in single throw of a dice $ = \dfrac{5}{6} $
Now, we need to find the probability of getting number $ 1 $ when a fair dice is thrown and it appears in even numbered throws.
Probability that number $ 1 $ appears in second throw $ = \dfrac{5}{6} \times \dfrac{1}{6} $
Probability that number $ 1 $ appears in fourth throw $ = \left( {\dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{1}{6}} \right) $
Probability that number $ 1 $ appears in sixth throw $ = \left( {\dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{1}{6}} \right) $
And so it goes on infinity
Thus, the probability of getting number 1 when a fair dice is thrown and it appears in even numbered throw $ = \left( {\dfrac{5}{6} \times \dfrac{1}{6}} \right) + \left( {\dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{1}{6}} \right) + \left( {\dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{1}{6}} \right) + - - - - $
The above expression is a Geometric progression of infinite terms where first term (a) $ = \dfrac{5}{{36}} $ and common ratio (r) $ = \dfrac{{25}}{{36}} $
So, Required probability $ = {S_\infty } $ \[ = \dfrac{{\dfrac{5}{{36}}}}{{1 - \dfrac{{25}}{{36}}}}\]
Simplifying the expression, we get,
\[ \Rightarrow {S_\infty }\]\[ = \dfrac{5}{{36 - 25}}\]
\[ \Rightarrow {S_\infty }\] $ = \dfrac{5}{{11}} $
Hence, probability of getting number $ 1 $ when a fair dice is thrown and it appears in an even numbered throw is $ \dfrac{5}{{11}} $ .
So, the correct answer is “ $ \dfrac{5}{{11}} $ ”.

Note: Such questions involving the concepts of probability and permutations and combinations are not so easy to deal with. Knowledge of the concepts of Permutation and combination and probability along with the calculation of factorials will be beneficial in solving more questions like this.