Question

A bag contains 12 two rupee coins, 7 one rupee coins and 4 half a rupee coins. If three coins are selected at random, then find the probability that
(i) the sum of three coins is maximum
(ii) the sum of three coins is minimum
(iii) each coin is of different value.

Answer 1

Hint : We will find the total sample available by adding all the number of coins in the bag. So, we get the sample space as ${}^{23}{{C}_{3}}$. For the first case, the sum of the coins is maximum when all the three coins drawn are of two rupee coins. For the second case the sum of the three coins will be minimum when all the drawn coins are of half a rupee coin. And for the third case, we will select one coin from each.

Complete step by step solution :
It is given in the question that a bag contains 12 two rupee coins, 7 one rupee coins and 4 half a rupee coins. If three coins are selected at random, then find the probability that (i) the sum of three coins is maximum (ii) the sum of three coins is minimum and (iii) each coin is of different value. We know that the formula for probability = $\dfrac{number\text{ }of\text{ }favourable\text{ }outcomes}{total\text{ }outcomes}$. We know that the total outcome will remain the same in all the three cases. The total outcome will be = ${}^{12+7+4}{{C}_{3}}={}^{23}{{C}_{3}}$. We know that the formula for ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. So, here, we have n = 23 and r = 3, so we get,
$\begin{aligned} & ^{23}{{C}_{3}}=\dfrac{23!}{3!\left( 23-3 \right)!} \\\ & {{\Rightarrow }^{23}}{{C}_{3}}=\dfrac{23\times 22\times 21\times 20!}{3!\times 20!} \\\ & {{\Rightarrow }^{23}}{{C}_{3}}=\dfrac{23\times 22\times 21}{3\times 2\times 1} \\\ & {{\Rightarrow }^{23}}{{C}_{3}}=\dfrac{10626}{6} \\\ & {{\Rightarrow }^{23}}{{C}_{3}}=1771 \\\ \end{aligned}$
Now we will find the probability for the first case.
(i) The sum of three coins is maximum.
So, for the maximum sum all the three coins drawn must be of two rupee coins. We have a total of 12 two rupee coins and 3 coins have to be drawn. So, we get the favoured outcome as = $^{12}{{C}_{3}}$. We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and we have here, n = 12 and r = 3. So, we will get,
$\begin{aligned} & ^{12}{{C}_{3}}=\dfrac{12!}{3!\left( 12-3 \right)!} \\\ & {{\Rightarrow }^{12}}{{C}_{3}}=\dfrac{12\times 11\times 10\times 9!}{3!\times 9!} \\\ & {{\Rightarrow }^{12}}{{C}_{3}}=\dfrac{12\times 11\times 10}{3\times 2\times 1} \\\ & {{\Rightarrow }^{12}}{{C}_{3}}=\dfrac{12\times 11\times 10}{6} \\\ & {{\Rightarrow }^{12}}{{C}_{3}}=2\times 11\times 10 \\\ & {{\Rightarrow }^{12}}{{C}_{3}}=220 \\\ \end{aligned}$
Thus, the probability that the sum of the three coins is maximum is equal to $\dfrac{220}{1771}\Rightarrow \dfrac{20}{161}$.
(ii) The sum of three coins is minimum.
So, we know the sum of the three coins is minimum when all the three coins drawn will be of half rupee coins. We have a total of 4 half rupee coins out of which 3 are drawn. So, the favoured outcome would be $^{4}{{C}_{3}}$. And we know that, ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. We have n = 4 and r = 3. So, we get,
$\begin{aligned} & ^{4}{{C}_{3}}=\dfrac{4!}{3!\left( 4-3 \right)!} \\\ & {{\Rightarrow }^{4}}{{C}_{3}}=\dfrac{4!}{3!\times 1!} \\\ & {{\Rightarrow }^{4}}{{C}_{3}}=\dfrac{4\times 3!}{1\times 3!} \\\ & {{\Rightarrow }^{4}}{{C}_{3}}=4 \\\ \end{aligned}$
Thus, the probability that the sum of the three coins is minimum will be equal to $\dfrac{4}{1771}$.
(iii) Each coin is of different value.
So, this means we have to draw each type of coin once. So, we have 12 two rupee coins, 7 one rupee coins and 4 half rupee coins. So, the favoured outcome would be = $^{12}{{C}_{1}}{{.}^{7}}{{C}_{1}}{{.}^{4}}{{C}_{1}}$. And we know that, ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. So, we will find the value for each of them. So, for 12 two rupee coins, we have,
$^{12}{{C}_{1}}=\dfrac{12!}{1!\left( 12-1 \right)!}\Rightarrow \dfrac{12!}{11!}\Rightarrow \dfrac{12\times 11!}{11!}\Rightarrow 12$
Similarly, we get for 7 one rupee coins as,
$^{7}{{C}_{1}}=\dfrac{7!}{1!\left( 7-1 \right)!}\Rightarrow \dfrac{7!}{6!}\Rightarrow \dfrac{7\times 6!}{6!}\Rightarrow 7$
And for, 4 half rupee coins we get,
$^{4}{{C}_{1}}=\dfrac{4!}{1!\left( 4-1 \right)!}\Rightarrow \dfrac{4!}{3!}\Rightarrow \dfrac{4\times 3!}{3!}\Rightarrow 4$
So, we get the favoured outcomes as $12\times 7\times 4=336$. Thus, the probability of drawing a coin of different value is equal to $\dfrac{336}{1771}$.

Note : Many students make a mistake while finding the probability of the third part of the solution. They may add the favoured probability as 12 + 7 + 4 = 23 and get the answer for the probability of each coin of different value as $\dfrac{23}{1771}$. We have to multiply the values $12\times 7\times 4$ and not add because when we draw $^{12}{{C}_{1}}$, the task is incomplete and it still remains incomplete when we draw $^{7}{{C}_{1}}$. The task is complete only when we draw $^{4}{{C}_{1}}$, so we have to multiply them and not add.