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# A single fair die is to be rolled until either a one or two is rolled. The probability that at least 3 rolls are needed, is (a) $\dfrac{2}{9}$  (b) $\dfrac{4}{5}$  (c) $\dfrac{5}{9}$  (d) $\dfrac{7}{9}$

Last updated date: 15th Aug 2024
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Hint: Here, we have to find the probability that at least 3 rolls are needed. The complement of this event would be the event that less than 2 rolls are needed. The probability of an event is equal to the probability of its complement event, subtracted from 1. We will find the probability of the complement event and subtract it from 1 to get the required answer.

Formula Used: The probability of an event is equal to the probability of its complement event, subtracted from 1.
The probability of an event is given by $P\left( E \right) = \dfrac{{{\rm{Number\, of\, favourable\, outcomes}}}}{{{\rm{Number\, of\, total\, outcomes}}}}$.

We will find the probability that at least 3 rolls are needed by subtracting the probability that 1 or 2 rolls are needed, from 1.
Let $X$ denote the number of rolls required to get a success, where getting a one or a two is a success.
We know that the possible outcomes when a die is rolled are 1, 2, 3, 4, 5, and 6.
Therefore, there are 6 possible outcomes.
The favourable outcomes are getting a 1 or a 2.
Therefore, the number of favourable outcomes is 2.
Thus, we get
Probability of getting a 1 or a 2 in one roll of a die $= \dfrac{2}{6} = \dfrac{1}{3}$
Thus, the probability of success is $\dfrac{1}{3}$.
Similarly, the unfavourable outcomes are getting a 3, 4, 5, or 6.
Therefore, the number of unfavourable outcomes is 4.
Thus, the probability of failure is $= \dfrac{4}{6} = \dfrac{2}{3}$.
Now, we know that the probability of an event is equal to the probability of its complement event, subtracted from 1.
We need to find the probability that at least 3 rolls are needed.
The complement of this event would be the event that less than 2 rolls are needed.
Therefore, we get
$P\left( {X \ge 3} \right) = 1 - P\left( {X < 2} \right)$
Rewriting the equation, we get
$\begin{array}{l} \Rightarrow P\left( {X \ge 3} \right) = 1 - \left[ {P\left( {X = 1} \right) + P\left( {X = 2} \right)} \right]\\ \Rightarrow P\left( {X \ge 3} \right) = 1 - P\left( {X = 1} \right) - P\left( {X = 2} \right)\end{array}$
The probability of getting a success in one roll of the die is $\dfrac{1}{3}$.
Thus, we get
$P\left( {X = 1} \right) = \dfrac{1}{3}$
The event that the success is obtained on the second roll of the die means that the first roll of the die was not a success.
Therefore, we get
$\begin{array}{l}P\left( {X = 2} \right) = P\left( {{\rm{failure}}} \right) \times P\left( {{\rm{success}}} \right)\\ \Rightarrow P\left( {X = 2} \right) = \dfrac{2}{3} \times \dfrac{1}{3}\\ \Rightarrow P\left( {X = 2} \right) = \dfrac{2}{9}\end{array}$
Substituting $P\left( {X = 1} \right) = \dfrac{1}{3}$ and $P\left( {X = 2} \right) = \dfrac{2}{9}$ in the equation $P\left( {X \ge 3} \right) = 1 - P\left( {X = 1} \right) - P\left( {X = 2} \right)$, we get
$\Rightarrow P\left( {X \ge 3} \right) = 1 - \dfrac{1}{3} - \dfrac{2}{9}$
Taking the L.C.M., we get
$\Rightarrow P\left( {X \ge 3} \right) = \dfrac{{9 - 3 - 2}}{9}$
Subtracting the terms in the numerator, we get
$P\left( {X \ge 3} \right) = \dfrac{4}{9}$
$\therefore$ The probability that at least 3 rolls are needed is $\dfrac{4}{9}$.

Note: None of the given options is correct.
We can also create a geometric progression with infinite terms to solve this question.
We can rewrite the probability that at least 3 rolls are needed as
$P\left( {X \ge 3} \right) = P\left( {X = 3} \right) + P\left( {X = 4} \right) + P\left( {X = 5} \right) + \ldots \ldots \ldots$
The event that the success is obtained on the third roll of the die means that the first two rolls of the die resulted in a failure.
Therefore, we get
$\begin{array}{l}P\left( {X = 3} \right) = P\left( {{\rm{failure}}} \right) \times P\left( {{\rm{failure}}} \right) \times P\left( {{\rm{success}}} \right)\\ \Rightarrow P\left( {X = 3} \right) = \dfrac{2}{3} \times \dfrac{2}{3} \times \dfrac{1}{3}\\ \Rightarrow P\left( {X = 3} \right) = \dfrac{4}{{27}}\end{array}$
Similarly, we get
$\begin{array}{l}P\left( {X = 4} \right) = P\left( {{\rm{failure}}} \right) \times P\left( {{\rm{failure}}} \right) \times P\left( {{\rm{failure}}} \right) \times P\left( {{\rm{success}}} \right)\\ \Rightarrow P\left( {X = 4} \right) = \dfrac{2}{3} \times \dfrac{2}{3} \times \dfrac{2}{3} \times \dfrac{1}{3}\\ \Rightarrow P\left( {X = 4} \right) = \dfrac{4}{{27}} \times \dfrac{2}{3}\end{array}$
$\begin{array}{l}P\left( {X = 5} \right) = P\left( {{\rm{failure}}} \right) \times P\left( {{\rm{failure}}} \right) \times P\left( {{\rm{failure}}} \right) \times P\left( {{\rm{failure}}} \right) \times P\left( {{\rm{success}}} \right)\\ \Rightarrow P\left( {X = 5} \right) = \dfrac{2}{3} \times \dfrac{2}{3} \times \dfrac{2}{3} \times \dfrac{2}{3} \times \dfrac{1}{3}\\ \Rightarrow P\left( {X = 5} \right) = \dfrac{4}{{27}} \times \dfrac{2}{3} \times \dfrac{2}{3}\end{array}$
Thus, we get
$\Rightarrow P\left( {X \ge 3} \right) = \dfrac{4}{{27}} + \dfrac{4}{{27}} \times \dfrac{2}{3} + \dfrac{4}{{27}} \times \dfrac{2}{3} \times \dfrac{2}{3} + \ldots \ldots \ldots$
We can observe that this is the sum of an geometric progression with infinite terms, where $a = \dfrac{4}{{27}}$ is the first term and $d = \dfrac{2}{3}$ is the common ratio.
The sum of a G.P. with infinite terms is given by the formula $S = \dfrac{a}{{1 - r}}$.
Therefore, we get
$\begin{array}{l} \Rightarrow \dfrac{4}{{27}} + \dfrac{4}{{27}} \times \dfrac{2}{3} + \dfrac{4}{{27}} \times \dfrac{2}{3} \times \dfrac{2}{3} + \ldots \ldots \ldots = \dfrac{{\dfrac{4}{{27}}}}{{1 - \dfrac{2}{3}}}\\ \Rightarrow P\left( {X \ge 3} \right) = \dfrac{{\dfrac{4}{{27}}}}{{1 - \dfrac{2}{3}}}\end{array}$
Simplifying the expression, we get
$\begin{array}{l} \Rightarrow P\left( {X \ge 3} \right) = \dfrac{{\dfrac{4}{{27}}}}{{\dfrac{1}{3}}}\\ \Rightarrow P\left( {X \ge 3} \right) = \dfrac{{4 \times 3}}{{1 \times 27}}\\ \Rightarrow P\left( {X \ge 3} \right) = \dfrac{4}{9}\end{array}$
$\therefore$ The probability that at least 3 rolls are needed is $\dfrac{4}{9}$.