A purse contains 4 coins; two coins having been drawn are found to be sovereigns. Find the chance that (1) all the coins are sovereigns (2) if the coins are replaced another drawing will give a sovereign
Answer
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Hint: We start solving the problem by finding the total possibilities of the sovereigns present in the purse and then find the probability of each of them. We then use the bayes theorem and find the probability that the purse has 4 sovereigns. Similarly, for problem (ii), we proceed in the backward of finding the probability of having sovereigns in the other two coins after the given coins are replaced. We then use the total probability theorem to get the required probability.
Complete step-by-step answer:
(i) According to the problem, we have a purse with four coins. If two coins are drawn and found to be sovereigns, we then need to find the probability that all the coins are sovereigns.
We solve this problem by using the Bayes theorem. We don’t know what is the total no. of sovereigns present in the purse. So, we first assume the probability of events ${{p}_{1}}$, ${{p}_{2}}$ and ${{p}_{3}}$ as discussed below.
${{p}_{1}}$- probability of the event of drawing two sovereigns if the purse has only two sovereigns.
${{p}_{2}}$- probability of the event of drawing two sovereigns if the purse has three sovereigns.
${{p}_{3}}$- probability of the event of drawing two sovereigns if the purse has three sovereigns.
Let us find the probabilities of events ${{p}_{1}}$, ${{p}_{2}}$, ${{p}_{3}}$.
So, we get ${{p}_{1}}=\dfrac{\text{total no}\text{. of ways to pick two sovereigns from a purse of 2 sovereigns}}{\text{total no}\text{. of ways to select two coins from a purse of 4 coins}}$.
We know that the no. of ways to select $r$ items out of $p$ items is ${}^{p}{{C}_{r}}$.
${{p}_{1}}=\dfrac{{}^{2}{{C}_{2}}}{{}^{4}{{C}_{2}}}$.
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and $n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ......\times 2\times 1$.
$\Rightarrow {{p}_{1}}=\dfrac{\dfrac{2!}{2!\left( 2-2 \right)!}}{\dfrac{4!}{2!\left( 4-2 \right)!}}$.
$\Rightarrow {{p}_{1}}=\dfrac{\dfrac{2!}{2!0!}}{\dfrac{4!}{2!2!}}$. We know that $0!=1$.
\[\Rightarrow {{p}_{1}}=\dfrac{\dfrac{1}{1}}{\dfrac{4\times 3}{2\times 1}}\].
\[\Rightarrow {{p}_{1}}=\dfrac{1}{6}\] ---(1).
Now, we have ${{p}_{2}}=\dfrac{\text{total no}\text{. of ways to pick two sovereigns from a purse of 3 sovereigns}}{\text{total no}\text{. of ways to select two coins from a purse of 4 coins}}$.
We know that the no. of ways to select $r$ items out of $p$ items is ${}^{p}{{C}_{r}}$.
${{p}_{2}}=\dfrac{{}^{3}{{C}_{2}}}{{}^{4}{{C}_{2}}}$.
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and $n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ......\times 2\times 1$.
$\Rightarrow {{p}_{2}}=\dfrac{\dfrac{3!}{2!\left( 3-2 \right)!}}{\dfrac{4!}{2!\left( 4-2 \right)!}}$.
$\Rightarrow {{p}_{2}}=\dfrac{\dfrac{3!}{2!1!}}{\dfrac{4!}{2!2!}}$.
\[\Rightarrow {{p}_{2}}=\dfrac{3}{\dfrac{4\times 3}{2\times 1}}\].
\[\Rightarrow {{p}_{2}}=\dfrac{1}{2}\] ---(2).
So, we have ${{p}_{3}}=\dfrac{\text{total no}\text{. of ways to pick two sovereigns from a purse of 4 sovereigns}}{\text{total no}\text{. of ways to select two coins from a purse of 4 coins}}$.
We know that the no. of ways to select $r$ items out of $p$ items is ${}^{p}{{C}_{r}}$.
${{p}_{3}}=\dfrac{{}^{4}{{C}_{2}}}{{}^{4}{{C}_{2}}}$.
\[\Rightarrow {{p}_{3}}=1\] ---(3).
Let us use the Bayes theorem to find the probability that the purse has 4 sovereigns if two of the drawn coins are sovereigns. Let us assume that probability is $s$.
So, we get $s=\dfrac{{{p}_{3}}}{{{p}_{1}}+{{p}_{2}}+{{p}_{3}}}$.
$\Rightarrow s=\dfrac{1}{\dfrac{1}{6}+\dfrac{1}{2}+1}$.
$\Rightarrow s=\dfrac{1}{\dfrac{1+3+6}{6}}$.
$\Rightarrow s=\dfrac{6}{10}$.
$\Rightarrow s=\dfrac{3}{5}$.
So, we have found the probability that all the coins in the purse are sovereign as $\dfrac{3}{5}$.
(ii) Now, we need to find the probability if the other drawn coin is sovereign if the drawn coins are replaced again in the purse.
Let us recall the possibilities of other two coins and assume those possibilities as ${{q}_{1}}$, ${{q}_{2}}$ and ${{q}_{3}}$.
Where ${{q}_{1}}$= the other two coins are sovereigns.
${{q}_{2}}$ = the other two coins contain one sovereign and one other.
${{q}_{3}}$ = the other two coins don't have any sovereign.
So, the probability of having one sovereign is equal to the probability of having other coins. This tells us that the probability of having one sovereign coin is $\dfrac{1}{2}$, and the probability of having another coin is $\dfrac{1}{2}$. Let us find the probabilities of the possibilities ${{q}_{1}}$, ${{q}_{2}}$ and ${{q}_{3}}$.
So, we have $P\left( {{q}_{1}} \right)=\dfrac{1}{2}\times \dfrac{1}{2}$.
$P\left( {{q}_{1}} \right)=\dfrac{1}{4}$ ---(4).
Now, we have $P\left( {{q}_{2}} \right)=\dfrac{1}{2}\times \dfrac{1}{2}\times 2$. Here two is multiplied because there are ways of having one sovereign and other i.e., ${{3}^{rd}}$ coin is sovereign and \[{{4}^{th}}\] is other, ${{3}^{rd}}$ coin is other and \[{{4}^{th}}\] is sovereign.
$P\left( {{q}_{2}} \right)=\dfrac{1}{2}$ ---(5).
So, we have $P\left( {{q}_{3}} \right)=\dfrac{1}{2}\times \dfrac{1}{2}$.
$P\left( {{q}_{3}} \right)=\dfrac{1}{4}$ ---(6).
Now, let us find the probability that a drawn coin is sovereign. Let us assume this event as ${{s}_{1}}$.
So, we have $P\left( {{s}_{1}} \right)=\left( P\left( {{q}_{1}} \right)\times P\left( {{d}_{1}} \right) \right)+\left( P\left( {{q}_{2}} \right)\times P\left( {{d}_{2}} \right) \right)+\left( P\left( {{q}_{3}} \right)\times P\left( {{d}_{3}} \right) \right)$ ---(7).
Where ${{d}_{1}}$-event of drawing two sovereigns if the purse has possibility ${{q}_{1}}$.
${{d}_{2}}$-event of drawing two sovereigns if the purse has possibility ${{q}_{2}}$.
${{d}_{3}}$-event of drawing two sovereigns if the purse has possibility ${{q}_{3}}$.
So, we get $P\left( {{d}_{1}} \right)=\dfrac{\text{total no}\text{. of ways to pick two sovereigns from a purse of 4 sovereigns}}{\text{total no}\text{. of ways to select two coins from a purse of 4 coins}}$.
$P\left( {{d}_{1}} \right)=\dfrac{{}^{4}{{C}_{2}}}{{}^{4}{{C}_{2}}}$.
\[P\left( {{d}_{1}} \right)=1\] ---(8).
Now, we have $P\left( {{d}_{2}} \right)=\dfrac{\text{total no}\text{. of ways to pick two sovereigns from a purse of 3 sovereigns}}{\text{total no}\text{. of ways to select two coins from a purse of 4 coins}}$.
$\Rightarrow P\left( {{d}_{2}} \right)=\dfrac{{}^{3}{{C}_{2}}}{{}^{4}{{C}_{2}}}$.
$\Rightarrow P\left( {{d}_{2}} \right)=\dfrac{\dfrac{3!}{2!\left( 3-2 \right)!}}{\dfrac{4!}{2!\left( 4-2 \right)!}}$.
$\Rightarrow P\left( {{d}_{2}} \right)=\dfrac{\dfrac{3!}{2!1!}}{\dfrac{4!}{2!2!}}$.
\[\Rightarrow P\left( {{d}_{2}} \right)=\dfrac{1}{2}\] ---(9).
So, we have $P\left( {{d}_{3}} \right)=\dfrac{\text{total no}\text{. of ways to pick two sovereigns from a purse of 4 sovereigns}}{\text{total no}\text{. of ways to select two coins from a purse of 4 coins}}$.
$\Rightarrow P\left( {{d}_{3}} \right)=\dfrac{{}^{2}{{C}_{2}}}{{}^{4}{{C}_{2}}}$.
$\Rightarrow P\left( {{d}_{3}} \right)=\dfrac{\dfrac{2!}{2!\left( 2-2 \right)!}}{\dfrac{4!}{2!\left( 4-2 \right)!}}$.
$\Rightarrow P\left( {{d}_{3}} \right)=\dfrac{\dfrac{2!}{2!0!}}{\dfrac{4!}{2!2!}}$.
\[\Rightarrow P\left( {{d}_{3}} \right)=\dfrac{1}{6}\] ---(10).
We substitute equations (4), (5), (6), (8), (9), (10) in equation (7).
So, we get $P\left( {{s}_{1}} \right)=\left( \dfrac{1}{4}\times 1 \right)+\left( \dfrac{1}{2}\times \dfrac{1}{2} \right)+\left( \dfrac{1}{4}\times \dfrac{1}{6} \right)$.
$\Rightarrow P\left( {{s}_{1}} \right)=\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{24}$.
$\Rightarrow P\left( {{s}_{1}} \right)=\dfrac{6+6+1}{24}$.
$\Rightarrow P\left( {{s}_{1}} \right)=\dfrac{13}{24}$.
So, we have found the probability that the other drawing is sovereign as $\dfrac{13}{24}$.
Note: Here we can see that both the problems given are proceeding in the same way but in the opposite direction. We should confuse when to use Bayes theorem and the total probability theorem while solving these problems. Whenever we get this type of problem, we first try to write all the possibilities and find the probabilities of them. We should not make calculation mistakes while solving this problem.
Complete step-by-step answer:
(i) According to the problem, we have a purse with four coins. If two coins are drawn and found to be sovereigns, we then need to find the probability that all the coins are sovereigns.
We solve this problem by using the Bayes theorem. We don’t know what is the total no. of sovereigns present in the purse. So, we first assume the probability of events ${{p}_{1}}$, ${{p}_{2}}$ and ${{p}_{3}}$ as discussed below.
${{p}_{1}}$- probability of the event of drawing two sovereigns if the purse has only two sovereigns.
${{p}_{2}}$- probability of the event of drawing two sovereigns if the purse has three sovereigns.
${{p}_{3}}$- probability of the event of drawing two sovereigns if the purse has three sovereigns.
Let us find the probabilities of events ${{p}_{1}}$, ${{p}_{2}}$, ${{p}_{3}}$.
So, we get ${{p}_{1}}=\dfrac{\text{total no}\text{. of ways to pick two sovereigns from a purse of 2 sovereigns}}{\text{total no}\text{. of ways to select two coins from a purse of 4 coins}}$.
We know that the no. of ways to select $r$ items out of $p$ items is ${}^{p}{{C}_{r}}$.
${{p}_{1}}=\dfrac{{}^{2}{{C}_{2}}}{{}^{4}{{C}_{2}}}$.
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and $n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ......\times 2\times 1$.
$\Rightarrow {{p}_{1}}=\dfrac{\dfrac{2!}{2!\left( 2-2 \right)!}}{\dfrac{4!}{2!\left( 4-2 \right)!}}$.
$\Rightarrow {{p}_{1}}=\dfrac{\dfrac{2!}{2!0!}}{\dfrac{4!}{2!2!}}$. We know that $0!=1$.
\[\Rightarrow {{p}_{1}}=\dfrac{\dfrac{1}{1}}{\dfrac{4\times 3}{2\times 1}}\].
\[\Rightarrow {{p}_{1}}=\dfrac{1}{6}\] ---(1).
Now, we have ${{p}_{2}}=\dfrac{\text{total no}\text{. of ways to pick two sovereigns from a purse of 3 sovereigns}}{\text{total no}\text{. of ways to select two coins from a purse of 4 coins}}$.
We know that the no. of ways to select $r$ items out of $p$ items is ${}^{p}{{C}_{r}}$.
${{p}_{2}}=\dfrac{{}^{3}{{C}_{2}}}{{}^{4}{{C}_{2}}}$.
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and $n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ......\times 2\times 1$.
$\Rightarrow {{p}_{2}}=\dfrac{\dfrac{3!}{2!\left( 3-2 \right)!}}{\dfrac{4!}{2!\left( 4-2 \right)!}}$.
$\Rightarrow {{p}_{2}}=\dfrac{\dfrac{3!}{2!1!}}{\dfrac{4!}{2!2!}}$.
\[\Rightarrow {{p}_{2}}=\dfrac{3}{\dfrac{4\times 3}{2\times 1}}\].
\[\Rightarrow {{p}_{2}}=\dfrac{1}{2}\] ---(2).
So, we have ${{p}_{3}}=\dfrac{\text{total no}\text{. of ways to pick two sovereigns from a purse of 4 sovereigns}}{\text{total no}\text{. of ways to select two coins from a purse of 4 coins}}$.
We know that the no. of ways to select $r$ items out of $p$ items is ${}^{p}{{C}_{r}}$.
${{p}_{3}}=\dfrac{{}^{4}{{C}_{2}}}{{}^{4}{{C}_{2}}}$.
\[\Rightarrow {{p}_{3}}=1\] ---(3).
Let us use the Bayes theorem to find the probability that the purse has 4 sovereigns if two of the drawn coins are sovereigns. Let us assume that probability is $s$.
So, we get $s=\dfrac{{{p}_{3}}}{{{p}_{1}}+{{p}_{2}}+{{p}_{3}}}$.
$\Rightarrow s=\dfrac{1}{\dfrac{1}{6}+\dfrac{1}{2}+1}$.
$\Rightarrow s=\dfrac{1}{\dfrac{1+3+6}{6}}$.
$\Rightarrow s=\dfrac{6}{10}$.
$\Rightarrow s=\dfrac{3}{5}$.
So, we have found the probability that all the coins in the purse are sovereign as $\dfrac{3}{5}$.
(ii) Now, we need to find the probability if the other drawn coin is sovereign if the drawn coins are replaced again in the purse.
Let us recall the possibilities of other two coins and assume those possibilities as ${{q}_{1}}$, ${{q}_{2}}$ and ${{q}_{3}}$.
Where ${{q}_{1}}$= the other two coins are sovereigns.
${{q}_{2}}$ = the other two coins contain one sovereign and one other.
${{q}_{3}}$ = the other two coins don't have any sovereign.
So, the probability of having one sovereign is equal to the probability of having other coins. This tells us that the probability of having one sovereign coin is $\dfrac{1}{2}$, and the probability of having another coin is $\dfrac{1}{2}$. Let us find the probabilities of the possibilities ${{q}_{1}}$, ${{q}_{2}}$ and ${{q}_{3}}$.
So, we have $P\left( {{q}_{1}} \right)=\dfrac{1}{2}\times \dfrac{1}{2}$.
$P\left( {{q}_{1}} \right)=\dfrac{1}{4}$ ---(4).
Now, we have $P\left( {{q}_{2}} \right)=\dfrac{1}{2}\times \dfrac{1}{2}\times 2$. Here two is multiplied because there are ways of having one sovereign and other i.e., ${{3}^{rd}}$ coin is sovereign and \[{{4}^{th}}\] is other, ${{3}^{rd}}$ coin is other and \[{{4}^{th}}\] is sovereign.
$P\left( {{q}_{2}} \right)=\dfrac{1}{2}$ ---(5).
So, we have $P\left( {{q}_{3}} \right)=\dfrac{1}{2}\times \dfrac{1}{2}$.
$P\left( {{q}_{3}} \right)=\dfrac{1}{4}$ ---(6).
Now, let us find the probability that a drawn coin is sovereign. Let us assume this event as ${{s}_{1}}$.
So, we have $P\left( {{s}_{1}} \right)=\left( P\left( {{q}_{1}} \right)\times P\left( {{d}_{1}} \right) \right)+\left( P\left( {{q}_{2}} \right)\times P\left( {{d}_{2}} \right) \right)+\left( P\left( {{q}_{3}} \right)\times P\left( {{d}_{3}} \right) \right)$ ---(7).
Where ${{d}_{1}}$-event of drawing two sovereigns if the purse has possibility ${{q}_{1}}$.
${{d}_{2}}$-event of drawing two sovereigns if the purse has possibility ${{q}_{2}}$.
${{d}_{3}}$-event of drawing two sovereigns if the purse has possibility ${{q}_{3}}$.
So, we get $P\left( {{d}_{1}} \right)=\dfrac{\text{total no}\text{. of ways to pick two sovereigns from a purse of 4 sovereigns}}{\text{total no}\text{. of ways to select two coins from a purse of 4 coins}}$.
$P\left( {{d}_{1}} \right)=\dfrac{{}^{4}{{C}_{2}}}{{}^{4}{{C}_{2}}}$.
\[P\left( {{d}_{1}} \right)=1\] ---(8).
Now, we have $P\left( {{d}_{2}} \right)=\dfrac{\text{total no}\text{. of ways to pick two sovereigns from a purse of 3 sovereigns}}{\text{total no}\text{. of ways to select two coins from a purse of 4 coins}}$.
$\Rightarrow P\left( {{d}_{2}} \right)=\dfrac{{}^{3}{{C}_{2}}}{{}^{4}{{C}_{2}}}$.
$\Rightarrow P\left( {{d}_{2}} \right)=\dfrac{\dfrac{3!}{2!\left( 3-2 \right)!}}{\dfrac{4!}{2!\left( 4-2 \right)!}}$.
$\Rightarrow P\left( {{d}_{2}} \right)=\dfrac{\dfrac{3!}{2!1!}}{\dfrac{4!}{2!2!}}$.
\[\Rightarrow P\left( {{d}_{2}} \right)=\dfrac{1}{2}\] ---(9).
So, we have $P\left( {{d}_{3}} \right)=\dfrac{\text{total no}\text{. of ways to pick two sovereigns from a purse of 4 sovereigns}}{\text{total no}\text{. of ways to select two coins from a purse of 4 coins}}$.
$\Rightarrow P\left( {{d}_{3}} \right)=\dfrac{{}^{2}{{C}_{2}}}{{}^{4}{{C}_{2}}}$.
$\Rightarrow P\left( {{d}_{3}} \right)=\dfrac{\dfrac{2!}{2!\left( 2-2 \right)!}}{\dfrac{4!}{2!\left( 4-2 \right)!}}$.
$\Rightarrow P\left( {{d}_{3}} \right)=\dfrac{\dfrac{2!}{2!0!}}{\dfrac{4!}{2!2!}}$.
\[\Rightarrow P\left( {{d}_{3}} \right)=\dfrac{1}{6}\] ---(10).
We substitute equations (4), (5), (6), (8), (9), (10) in equation (7).
So, we get $P\left( {{s}_{1}} \right)=\left( \dfrac{1}{4}\times 1 \right)+\left( \dfrac{1}{2}\times \dfrac{1}{2} \right)+\left( \dfrac{1}{4}\times \dfrac{1}{6} \right)$.
$\Rightarrow P\left( {{s}_{1}} \right)=\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{24}$.
$\Rightarrow P\left( {{s}_{1}} \right)=\dfrac{6+6+1}{24}$.
$\Rightarrow P\left( {{s}_{1}} \right)=\dfrac{13}{24}$.
So, we have found the probability that the other drawing is sovereign as $\dfrac{13}{24}$.
Note: Here we can see that both the problems given are proceeding in the same way but in the opposite direction. We should confuse when to use Bayes theorem and the total probability theorem while solving these problems. Whenever we get this type of problem, we first try to write all the possibilities and find the probabilities of them. We should not make calculation mistakes while solving this problem.
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