Question

A fair dice is rolled. The probability that the first time \[1\] occurs at the even throw is
A. \[\dfrac{1}{6}\]
B. \[\dfrac{5}{{11}}\]
C. \[\dfrac{6}{{11}}\]
D. \[\dfrac{5}{{36}}\]

Answer 1

Hint: Probability means possibility of happening of an event . It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one. Probability of an event can never be negative.

Complete step by step answer:
The meaning of probability is basically the extent to which something is likely to happen. This is the basic probability theory, which is also used in the probability distribution, where you will learn the probability of outcomes for a random experiment. To find the probability of a single event to occur, first, we should know the total number of possible outcomes.
We know that one die has six outcomes such as \[1\] , \[2\] , \[3\] , \[4\] , \[5\] and \[6\] .
We know that Probability (event) \[ = \dfrac{{Number\;of\;favourable\;outcomes}}{{Total\;n umber\;of\;outcomes}}\]
Sum of an infinite Geometric Progression (GP)
 \[ = \dfrac{{{a_1}}}{{1 - r}}\]
 Where \[{a_1}\] is the first term and \[r\] is the common ratio .
Let E be the event of getting \[1\] on a dice .
Therefore we know that
\[P(E) = \dfrac{1}{6}\] and \[P(\overline E ) = \dfrac{5}{6}\]
Therefore P(first time \[1\] occurs at the even throw) \[ = {t_2}\] or \[{t_4}\] or \[{t_6}\] or \[{t_8}\] … and so on
 P(first time \[1\] occurs at the even throw) \[ = \left\{ {P(\overline E )P(E)} \right\} + \left\{ {P(\overline E )P(\overline E )P(\overline E )P(E)} \right\} + ...\infty \]
 \[ = \left( {\dfrac{5}{6}} \right)\left( {\dfrac{1}{6}} \right) + {\left( {\dfrac{5}{6}} \right)^3}\left( {\dfrac{1}{6}} \right) + {\left( {\dfrac{5}{6}} \right)^5}\left( {\dfrac{1}{6}} \right) + ...\infty \]
This becomes an infinite Geometric Progression (GP) with first term as \[\dfrac{5}{{36}}\] and common ratio as \[\dfrac{{25}}{{36}}\]
Therefore P(first time \[1\] occurs at the even throw) \[ = \left( {\dfrac{{\dfrac{5}{{36}}}}{{1 - \dfrac{{25}}{{36}}}}} \right)\]
 \[ = \left( {\dfrac{{\dfrac{5}{{36}}}}{{\dfrac{{36 - 25}}{{36}}}}} \right)\]
Which simplifies to \[ = \left( {\dfrac{{\dfrac{5}{{36}}}}{{\dfrac{{11}}{{36}}}}} \right)\]
Therefore P(first time \[1\] occurs at the even throw) \[ = \dfrac{5}{{11}}\]

So, the correct answer is “Option B”.

Note: The meaning of probability is basically the extent to which something is likely to happen. Probability of any event can be between \[0\] and \[1\] only. Probability of any event can never be greater than \[1\] . Probability of any event can never be negative.