Question

A box contains $4$ Choco bars and $4$ ice creams. Tom eats $3$ of them, by randomly choosing. What is the probability of choosing $2$ Choco bars and $1$ ice-cream?

Answer 1

First we have to find the total number of possible outcomes and the total number of required outcomes. After finding the probability for all the required outcomes, we add them to get the final probability.

Formula used:
Probability is a type of ratio where we compare how many times an outcome can occur compared to all possible outcomes.
${\text{Probability = }}\dfrac{{{\text{The number of wanted outcomes}}}}{{{\text{The number of possible outcomes}}}}$

Complete step-by-step answer:
It is given that the box contains $4$ Choco bars and $4$ ice creams.
Also, Tom eats $3$ of them, by randomly choosing.
Assuming the set containing C = chocolate, and I = Ice cream,
We get, as the initial set \left\\{ {I,I,I,I,C,C,C,C} \right\\}.
Acceptable outcomes are \left\\{ {I,C,C} \right\\},{\text{ }}\left\\{ {C,I,C} \right\\},{\text{ and}}\left\\{ {C,C,I} \right\\} .
First round:
Total set is \left\\{ {I,I,I,I,C,C,C,C} \right\\}
So the total possible outcomes is $^8{C_1}$
The possible cases for choosing either a chocolate or an ice cream is $^4{C_1}$(since we are choosing only one from either four chocolates or $4$ ice creams).
The probability of choosing either chocolate or ice cream in the first random pickup is
$= \dfrac{{^4{C_1}}}{{^8{C_1}}} = \dfrac{{\dfrac{{4!}}{{1!3!}}}}{{\dfrac{{8!}}{{1!7!}}}} = \dfrac{{\dfrac{{4.3!}}{{3!}}}}{{\dfrac{{8.7!}}{{7!}}}} = \dfrac{4}{8} = \dfrac{1}{2}$
Second round:
One random pickup is already done so the possible outcome is $^7{C_1}$
When Tom get chocolate if first pick was ice cream, the set is \left\\{ {I,I,I,C,C,C,C} \right\\}
So probability to get chocolate if first pick was ice cream is
$= \dfrac{{^4{C_1}}}{{^7{C_1}}} = \dfrac{{\dfrac{{4!}}{{1!3!}}}}{{\dfrac{{7!}}{{1!6!}}}} = \dfrac{{\dfrac{{4.3!}}{{3!}}}}{{\dfrac{{7.6!}}{{6!}}}} = \dfrac{4}{7}$
When Tom get chocolate if first pick was chocolate, the set is \left\\{ {I,I,I,I,C,C,C} \right\\}
So probability to get chocolate if first pick was chocolate is
$= \dfrac{{^3{C_1}}}{{^7{C_1}}} = \dfrac{{\dfrac{{3!}}{{1!2!}}}}{{\dfrac{{7!}}{{1!6!}}}} = \dfrac{{\dfrac{{3.2!}}{{2!}}}}{{\dfrac{{7.6!}}{{6!}}}} = \dfrac{3}{7}$
When Tom get ice cream if first pick was chocolate, the set is\;\left\\{ {I,I,I,I,C,C,C} \right\\}
So probability to get ice cream if first pick was chocolate is
$= \dfrac{{^4{C_1}}}{{^7{C_1}}} = \dfrac{{\dfrac{{4!}}{{1!3!}}}}{{\dfrac{{7!}}{{1!6!}}}} = \dfrac{{\dfrac{{4.3!}}{{3!}}}}{{\dfrac{{7.6!}}{{6!}}}} = \dfrac{4}{7}$
Third round:
Two random pickup is already done so the possible outcome is $^6{C_1}$
When Tom pick chocolate if previous picks yielded one of each, the set is\left\\{ {I,I,I,C,C,C} \right\\}
So probability to pick chocolate if previous picks yielded one of each
$= \dfrac{{^3{C_1}}}{{^6{C_1}}} = \dfrac{{\dfrac{{3!}}{{1!2!}}}}{{\dfrac{{6!}}{{1!5!}}}} = \dfrac{{\dfrac{{3.2!}}{{2!}}}}{{\dfrac{{6.5!}}{{5!}}}} = \dfrac{3}{6} = \dfrac{1}{2}$
When Tom pick ice cream if first pick and second picks were chocolate, the set is \left\\{ {I,I,I,I,C,C} \right\\}
So probability to pick ice cream if first pick and second picks were chocolate,
$= \dfrac{{^4{C_1}}}{{^6{C_1}}} = \dfrac{{\dfrac{{4!}}{{1!3!}}}}{{\dfrac{{6!}}{{1!5!}}}} = \dfrac{{\dfrac{{4.3!}}{{3!}}}}{{\dfrac{{6.5!}}{{5!}}}} = \dfrac{4}{6} = \dfrac{2}{3}$
For \left\\{ {C,C,I} \right\\} we have the probability =$\dfrac{1}{2} \times \dfrac{3}{7} \times \dfrac{2}{3} = \dfrac{1}{7}$.
For \left\\{ {C,I,C} \right\\}{\text{ and }}\left\\{ {I,C,C} \right\\} the probability is the same $\dfrac{1}{2} \times \dfrac{4}{7} \times \dfrac{1}{2} = \dfrac{1}{7}$.
In total we thus get,
$\Rightarrow \dfrac{1}{7} + \dfrac{1}{7} + \dfrac{1}{7} = \dfrac{3}{7}$
Thus, the probability of choosing $2$ Choco bars and $1$ ice-cream is $\dfrac{3}{7}$.

Note: A combination is a grouping or subset of items.
For a combination,
$C\left( {n,r} \right){ = ^n}{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$
Where, factorial $n$ is denoted by $n!$ and defined by
$n! = n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 4} \right) \ldots \ldots .2.1$