Question

3 mangoes and 3 apples are kept in a box. If 2 fruits are chosen at random, find the probability that one is a mango and the other is an apple is:
$\left( A \right){\text{ }}\dfrac{3}{5}$
$\left( B \right){\text{ }}\dfrac{5}{6}$
$\left( C \right){\text{ }}\dfrac{1}{{36}}$
$\left( D \right){\text{ }}\dfrac{7}{{36}}$

Answer 1

Hint: Here in this question we have to choose and we know that if we have to choose then we always use the combination. So here we have a total of $6$ fruits out of which we have to choose $2$ . And to find the probability we also need favorable ways. Then we can calculate the probability.

Formula used:
Combination,
${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Here, $n$, will be the number of items in the set
$r$, will be the number of items chosen from the set
Probability,
$P\left( A \right) = \dfrac{{Number{\text{ of favourable outcome}}}}{{Total{\text{ number of favourable outcomes}}}}$
Here, $P(A)$ , will be the probability of any event named $A$

Complete step by step solution:
Here, in this question, we will first find the total number of ways
Therefore, the total number of ways will be
$ \Rightarrow n(s) = {}^6{C_2}$
And on solving this combination by using the formula, we get
$ \Rightarrow n(s) = \dfrac{{6!}}{{2!\left( {6 - 2} \right)!}}$
And on solving the braces part of it, we get
$ \Rightarrow n(s) = \dfrac{{6!}}{{2! \times 4!}}$
Now on expanding and canceling the like term from the equation, we get
$ \Rightarrow n(s) = \dfrac{{6 \times 5}}{{2 \times 1}}$
And on solving it, we get
$ \Rightarrow n(s) = 15$
So the favorable number of ways will be
$ \Rightarrow n(E) = {}^3{C_1} \times {}^3{C_1}$
And on solving this combination by using the formula, we get
$ \Rightarrow n(E) = \dfrac{{3!}}{{1!\left( {3 - 1} \right)!}} \times \dfrac{{3!}}{{1!\left( {3 - 1} \right)!}}$
Now on expanding and canceling the like term from the equation, we get
\[ \Rightarrow n(E) = \dfrac{{3 \times 2 \times 1}}{{1 \times 2 \times 1}} \times \dfrac{{3 \times 2 \times 1}}{{1 \times 2 \times 1}}\]
And on solving it, we get
$ \Rightarrow n(E) = 3 \times 3$
And solving the multiplication, we get
$ \Rightarrow n(E) = 9$
Therefore, the required probability will be
$ \Rightarrow P\left( A \right) = \dfrac{{n(E)}}{{n(S)}}$
On substituting the values, we get
$ \Rightarrow P\left( A \right) = \dfrac{9}{{15}}$
And on solving it, we get
$ \Rightarrow P\left( A \right) = \dfrac{3}{5}$
Therefore, the probability will be $\dfrac{3}{5}$ .

Hence, the option $\left( a \right)$ will be correct.

Note: For solving this type of question the main thing is we have to know where to use permutation and where to use a combination. So whenever there is an arrangement then we choose permutation and for combination, it will be a combination of the ways. So by using this we can answer this type of question.