Question

A box contains 10 red balls and 15 green balls. If two balls are drawn in succession then the probability that one is red and other is green is:
A. $\dfrac{1}{3}$
B. $\dfrac{1}{2}$
C. $\dfrac{1}{4}$
D. None of these

Answer 1

Here the given question is of probability in which the solution needs the probability of the given condition, which is two balls are drawn in succession, here we need to use the property of combination, which says the total possible outcome for any condition like we are provided with. And there we will calculate total outcome and favorable outcome.
Formulae Used: The formulae for combination:
${ \Rightarrow ^n}{C_m} = \dfrac{{n!}}{{\left( {n - m} \right)! \times m!}}$
$\Rightarrow probability = \dfrac{{favourable\,outcome}}{{total\,outcome}}$

Complete step by step answer:
Here in the above question we need to calculate probability for the given condition, and we know that for calculating the probability for an given condition, we need favorable outcome and total outcome, and for that we will use combination property here, on solving we get:
Total outcome for the given condition that is two balls are drawn from the total number of balls, we have:

$${ \Rightarrow ^{25}}{C_2} = \dfrac{{25!}}{{\left( {25 - 2} \right)! \times 2!}} \\\ { \Rightarrow ^{25}}{C_2} = \dfrac{{25 \times 24 \times 23!}}{{23! \times (2 \times 1)}} = \dfrac{{600}}{2} = 300 \\\$$

Here, the total number of balls is twenty five, and balls drawn are two.
Now we have to calculate the outcome of one red ball and one green ball drawn, is:
One red ball drawn out of ten red balls we have:
${ \Rightarrow ^{10}}{C_1}$

One green ball drawn out of fifteen green balls we have:
${ \Rightarrow ^{15}}{C_1}$
Here both condition are acting simultaneously, hence both outcome will be multiplied, collectively we have:
Favorable outcome:

$${ \Rightarrow ^{15}}{C_1}{ \times ^{15}}{C_1} = \dfrac{{10!}}{{\left( {10 - 1} \right)! \times 1!}} \times \dfrac{{15!}}{{\left( {15 - 1} \right)! \times 1!}} \\\ { \Rightarrow ^{15}}{C_1}{ \times ^{15}}{C_1} = \dfrac{{10 \times 9!}}{{9!}} \times \dfrac{{15 \times 14!}}{{14!}} = 10 \times 15 = 150 \\\$$

Now probability:
$\Rightarrow probability = \dfrac{{favourable\,outcome}}{{total\,outcome}} = \dfrac{{150}}{{300}} = \dfrac{1}{2}$

So, the correct answer is “Option B”.

Note: Here the given question we need to use properties of permutation and combination property, this property helps to find the combination of a particular condition, which can arrive out of several possibilities and is used while solving probability problems.