$A$ and $B$ throw a dice alternatively till one of them gets a $6$ and wins the game. Find their respective probabilities of winning if $A$ starts first.
Last updated date: 18th Mar 2023
•
Total views: 307.2k
•
Views today: 4.03k
Answer
307.2k+ views
Hint: Try to find out probability in all those possible cases in which $A$ wins and then take the union of all these cases by simply adding these probabilities. To add the probabilities, use a formula for the sum of infinite G.P.
In the question, it is given that a person wins if he gets $6$ on the dice. A dice has a total
of six sides of which one of the sides has $6$ on it. So, probability of getting $6$ is,
$P\left( 6 \right)=\dfrac{1}{6}$
Let us consider $X$ as an event which denotes getting $6$ on the dice and $Y$ as an event for not
getting $6$ on the dice.
$\Rightarrow P\left( X \right)=\dfrac{1}{6}$
And $P\left( Y \right)=\dfrac{5}{6}$
To find the probability that $A$ wins, let us make all the possible cases.
Case 1: $A$ wins at first throw of dice
$\begin{align}
& P\left( A-wins \right)=P\left( X \right) \\
& \Rightarrow P\left( A-wins \right)=\dfrac{1}{6} \\
\end{align}$
Since $A$ and $B$ throw the dice alternative, so now, $B$ will throw dice. Now for the third throw,
$A$ will throw the dice.
Case 2: $A$ wins at the third throw of dice
$\begin{align}
& P\left( A\text{ }wins \right)=P\left( no\text{ }six\text{ }by\text{ }A \right).P\left( no\text{ }six\text{
}by\text{ B} \right).P\left( six\text{ }by\text{ }A \right) \\
& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{5}{6}.\dfrac{5}{6}.\dfrac{1}{6} \\
& \Rightarrow P\left( A\text{ }wins \right)={{\left( \dfrac{5}{6} \right)}^{2}}.\dfrac{1}{6} \\
\end{align}$
Now $B$ will throw on the fourth throw. Then again, $A$ will throw on the fifth throw of the dice.
Case 3: $A$ wins at the fifth throw
$\begin{align}
& P\left( A\text{ }wins \right)=P\left( no\text{ }six\text{ }by\text{ }A \right).P\left( no\text{ }six\text{
}by\text{ B} \right).P\left( no\text{ }six\text{ }by\text{ }A \right).P\left( no\text{ }six\text{ }by\text{ B}
\right).P\left( six\text{ }by\text{ }A \right) \\
& \Rightarrow P\left( A\text{ }wins
\right)=\dfrac{5}{6}.\dfrac{5}{6}.\dfrac{5}{6}.\dfrac{5}{6}.\dfrac{1}{6} \\
& \Rightarrow P\left( A\text{ }wins \right)={{\left( \dfrac{5}{6} \right)}^{4}}.\dfrac{1}{6} \\
\end{align}$
Similarly, we can find the probability of $A$ winning on the odd number of throws. The probability
of $A$ winning at ${{n}^{th}}$ thrown is equal to ${{\left( \dfrac{5}{6} \right)}^{n-1}}.\dfrac{1}{6}$ .
Since all the above listed cases are possible, the probability of $A$ winning is given by the union of
all these cases i.e. we have to add all these cases. Hence, the probability that $A$ wins is,
\[\begin{align}
& P\left( A\text{ }wins \right)=\dfrac{1}{6}+{{\left( \dfrac{5}{6} \right)}^{2}}\dfrac{1}{6}+{{\left(
\dfrac{5}{6} \right)}^{4}}\dfrac{1}{6}+{{\left( \dfrac{5}{6} \right)}^{6}}\dfrac{1}{6}+......... \\
& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{1}{6}\left( 1+{{\left( \dfrac{5}{6}
\right)}^{2}}+{{\left( \dfrac{5}{6} \right)}^{4}}+{{\left( \dfrac{5}{6} \right)}^{6}}+.........
\right)............\left( 1 \right) \\
\end{align}\]
The above term is an infinite G.P. Since the common ratio of the above G.P. if \[{{\left( \dfrac{5}{6}
\right)}^{2}}\] which is less than $1$, we can find it’s sum.
For an infinite G.P. $a,ar,a{{r}^{2}},a{{r}^{3}}.......$ the sum is given by,’
${{S}_{\infty }}=\dfrac{a}{1-r}...........\left( 2 \right)$
Using formula $\left( 2 \right)$ to find the sum of infinite GP in equation $\left( 1 \right)$ with
$a=1,r={{\left( \dfrac{5}{6} \right)}^{2}}$, we get,
\[\begin{align}
& P\left( A\text{ }wins \right)=\dfrac{1}{6}\left( \dfrac{1}{1-{{\left( \dfrac{5}{6} \right)}^{2}}} \right)
\\
& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{1}{6}\left( \dfrac{1}{1-\dfrac{25}{36}} \right) \\
& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{1}{6}\left( \dfrac{1}{\dfrac{11}{36}} \right) \\
& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{1}{6}\left( \dfrac{36}{11} \right) \\
& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{6}{11}...........\left( 3 \right) \\
\end{align}\]
In probability, we have a property,
$P\left( event \right)+P\left( complement\text{ }of\text{ }the\text{ }event \right)=1.......\left( 4
\right)$
Since complement of $A$ wins is $B$ wins, from $\left( 4 \right)$, we have,
$\begin{align}
& P\left( A\text{ }wins \right)+P\left( \text{B }wins \right)=1 \\
& \Rightarrow P\left( \text{B }wins \right)=1-P\left( A\text{ }wins \right) \\
\end{align}$
From $\left( 3 \right)$ substituting \[P\left( A\text{ }wins \right)=\dfrac{6}{11}\] in the above
equation, we get,
$\begin{align}
& P\left( \text{B }wins \right)=1-\dfrac{6}{11} \\
& \Rightarrow P\left( \text{B }wins \right)=\dfrac{5}{11} \\
\end{align}$
Note: One must know that the formula for the sum of infinite G.P. is applicable only for those G.P. in which the common ratio is less than $1$. If the common ratio is greater than $1$, we cannot use that formula to calculate sum.
In the question, it is given that a person wins if he gets $6$ on the dice. A dice has a total
of six sides of which one of the sides has $6$ on it. So, probability of getting $6$ is,
$P\left( 6 \right)=\dfrac{1}{6}$
Let us consider $X$ as an event which denotes getting $6$ on the dice and $Y$ as an event for not
getting $6$ on the dice.
$\Rightarrow P\left( X \right)=\dfrac{1}{6}$
And $P\left( Y \right)=\dfrac{5}{6}$
To find the probability that $A$ wins, let us make all the possible cases.
Case 1: $A$ wins at first throw of dice
$\begin{align}
& P\left( A-wins \right)=P\left( X \right) \\
& \Rightarrow P\left( A-wins \right)=\dfrac{1}{6} \\
\end{align}$
Since $A$ and $B$ throw the dice alternative, so now, $B$ will throw dice. Now for the third throw,
$A$ will throw the dice.
Case 2: $A$ wins at the third throw of dice
$\begin{align}
& P\left( A\text{ }wins \right)=P\left( no\text{ }six\text{ }by\text{ }A \right).P\left( no\text{ }six\text{
}by\text{ B} \right).P\left( six\text{ }by\text{ }A \right) \\
& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{5}{6}.\dfrac{5}{6}.\dfrac{1}{6} \\
& \Rightarrow P\left( A\text{ }wins \right)={{\left( \dfrac{5}{6} \right)}^{2}}.\dfrac{1}{6} \\
\end{align}$
Now $B$ will throw on the fourth throw. Then again, $A$ will throw on the fifth throw of the dice.
Case 3: $A$ wins at the fifth throw
$\begin{align}
& P\left( A\text{ }wins \right)=P\left( no\text{ }six\text{ }by\text{ }A \right).P\left( no\text{ }six\text{
}by\text{ B} \right).P\left( no\text{ }six\text{ }by\text{ }A \right).P\left( no\text{ }six\text{ }by\text{ B}
\right).P\left( six\text{ }by\text{ }A \right) \\
& \Rightarrow P\left( A\text{ }wins
\right)=\dfrac{5}{6}.\dfrac{5}{6}.\dfrac{5}{6}.\dfrac{5}{6}.\dfrac{1}{6} \\
& \Rightarrow P\left( A\text{ }wins \right)={{\left( \dfrac{5}{6} \right)}^{4}}.\dfrac{1}{6} \\
\end{align}$
Similarly, we can find the probability of $A$ winning on the odd number of throws. The probability
of $A$ winning at ${{n}^{th}}$ thrown is equal to ${{\left( \dfrac{5}{6} \right)}^{n-1}}.\dfrac{1}{6}$ .
Since all the above listed cases are possible, the probability of $A$ winning is given by the union of
all these cases i.e. we have to add all these cases. Hence, the probability that $A$ wins is,
\[\begin{align}
& P\left( A\text{ }wins \right)=\dfrac{1}{6}+{{\left( \dfrac{5}{6} \right)}^{2}}\dfrac{1}{6}+{{\left(
\dfrac{5}{6} \right)}^{4}}\dfrac{1}{6}+{{\left( \dfrac{5}{6} \right)}^{6}}\dfrac{1}{6}+......... \\
& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{1}{6}\left( 1+{{\left( \dfrac{5}{6}
\right)}^{2}}+{{\left( \dfrac{5}{6} \right)}^{4}}+{{\left( \dfrac{5}{6} \right)}^{6}}+.........
\right)............\left( 1 \right) \\
\end{align}\]
The above term is an infinite G.P. Since the common ratio of the above G.P. if \[{{\left( \dfrac{5}{6}
\right)}^{2}}\] which is less than $1$, we can find it’s sum.
For an infinite G.P. $a,ar,a{{r}^{2}},a{{r}^{3}}.......$ the sum is given by,’
${{S}_{\infty }}=\dfrac{a}{1-r}...........\left( 2 \right)$
Using formula $\left( 2 \right)$ to find the sum of infinite GP in equation $\left( 1 \right)$ with
$a=1,r={{\left( \dfrac{5}{6} \right)}^{2}}$, we get,
\[\begin{align}
& P\left( A\text{ }wins \right)=\dfrac{1}{6}\left( \dfrac{1}{1-{{\left( \dfrac{5}{6} \right)}^{2}}} \right)
\\
& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{1}{6}\left( \dfrac{1}{1-\dfrac{25}{36}} \right) \\
& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{1}{6}\left( \dfrac{1}{\dfrac{11}{36}} \right) \\
& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{1}{6}\left( \dfrac{36}{11} \right) \\
& \Rightarrow P\left( A\text{ }wins \right)=\dfrac{6}{11}...........\left( 3 \right) \\
\end{align}\]
In probability, we have a property,
$P\left( event \right)+P\left( complement\text{ }of\text{ }the\text{ }event \right)=1.......\left( 4
\right)$
Since complement of $A$ wins is $B$ wins, from $\left( 4 \right)$, we have,
$\begin{align}
& P\left( A\text{ }wins \right)+P\left( \text{B }wins \right)=1 \\
& \Rightarrow P\left( \text{B }wins \right)=1-P\left( A\text{ }wins \right) \\
\end{align}$
From $\left( 3 \right)$ substituting \[P\left( A\text{ }wins \right)=\dfrac{6}{11}\] in the above
equation, we get,
$\begin{align}
& P\left( \text{B }wins \right)=1-\dfrac{6}{11} \\
& \Rightarrow P\left( \text{B }wins \right)=\dfrac{5}{11} \\
\end{align}$
Note: One must know that the formula for the sum of infinite G.P. is applicable only for those G.P. in which the common ratio is less than $1$. If the common ratio is greater than $1$, we cannot use that formula to calculate sum.
Recently Updated Pages
Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts
Ray optics is valid when characteristic dimensions class 12 physics CBSE

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?

Alfred Wallace worked in A Galapagos Island B Australian class 12 biology CBSE

Imagine an atom made up of a proton and a hypothetical class 12 chemistry CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

How do you define least count for Vernier Calipers class 12 physics CBSE

Why is the cell called the structural and functional class 12 biology CBSE

A 30 solution of H2O2 is marketed as 100 volume hydrogen class 11 chemistry JEE_Main
