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# A box contains $12$ two rupee coins, $7$ one rupee coins and $4$ half rupee coins. If $3$ are selected at random, find the probability thatA) Sum of three coins is maximum.B) Each coin is of different value.C) Selection contains at least one-rupee coinD) All selected three coins have the same value.

Last updated date: 20th Jun 2024
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Hint: Firstly, note down the number of different kinds of coins we have. Use combinations to calculate the number of combinations possible for each event mentioned in the question. Divide it with the number of all possible combinations for the given question.

Given that,
A box contains $12$ two rupee coins, $7$ one rupee coins and $4$ half rupee coins.
Total number of coins $= 12 + 7 + 4$
$= 23$
Out of all $3$ coins were selected. Now we are asked to solve this problem in four sub parts.
The first one is to find the probability of sum of three coins is maximum,
In order to get the maximum, we need to select the coins with high value.
Among all the coins two-rupee coins are with high value. So, all the three coins must be two-rupee coins.
Now selecting $n$ items from $m$ items can be simply found by the formula $^m{C_n}$.
And the value $^m{C_n}$ can be calculated as $\dfrac{{m!}}{{n!\left( {m - n} \right)!}}$
Before calculating remaining stuff, firstly let us calculate the number of ways in which we can select three coins from the box and that is $^{23}{C_3}$
Where $^{23}{C_3} = \dfrac{{23!}}{{3!\left( {23 - 3} \right)!}}$
$= 1771$
So, selecting three from twelve two-rupee coins ${ = ^{12}}{C_3}$
$= \dfrac{{12!}}{{3!\left( {12 - 3} \right)!}} \\ = \dfrac{{12!}}{{3!\left( 9 \right)!}} \\ = \dfrac{{12 \times 11 \times 10 \times 9!}}{{3 \times 2 \times 1 \times \left( 9 \right)!}} \\ = 220 \\$
So, the required probability is $\dfrac{{220}}{{1771}} = \dfrac{{20}}{{161}}$.
Now, the second one,
To get each coin with different value we need to take one from each group and that is as follows:
Number of selections with all different values ${ = ^{12}}{C_1}{ \times ^7}{C_1}{ \times ^4}{C_1}$
$= 12 \times 7 \times 4 \\ = 336 \\$
Required probability for the coins with different values is $\dfrac{{336}}{{1771}} = \dfrac{{48}}{{253}}$
In the third case,
Selection must contain at least one-rupee coin and that is as follows:
The combination of one one-rupee coin and remaining coins or the combination of two one-rupee coins and remaining coins or all the three coins are one-rupee coins.
${ \Rightarrow ^7}{C_1}{ \times ^{16}}{C_2}{ + ^7}{C_2}{ \times ^{16}}{C_1}{ + ^7}{C_3}{ \times ^{16}}{C_0}$$= 7 \times 120 + 21 \times 16 + 35$
$= 840 + 336 + 35 \\ = 1211 \\$
So, the required probability will be $\dfrac{{1211}}{{1771}} = 0.68$
In the fourth case all the selected coins should have the same value.
It means that we should select from only one set as follows:
${ \Rightarrow ^{12}}{C_3}{ + ^7}{C_3}{ + ^4}{C_3} = 220 + 35 + 4$
$= 259$
So, he required probability is $\dfrac{{259}}{{1771}} = 0.14$

Note: Probability is defined as the possible number of outcomes to the total number of outcomes. Permutations is the arrangement of the items which are already present in the set. Combination is the selection of a few items from a set of many items.