Question

A bag contains 3 red balls and 7 black balls. Two balls are drawn one by one at a time at random without replacement. If second drawn ball is red then what is the probability that first drawn ball is also red.

Answer 1

Hint: In this question, we need to first list all the possibilities or the outcomes of the first ball characteristics when the second ball is red and then use the principle of conditional probability to find the probability of first ball being red when second ball drawn is red.

Complete step by step answer:
Here in this question, we have 3 red balls and 7 black balls in a bag which means we have total number of balls in the bag. Therefore, the total number of balls is n(s) = 10. It is given when two balls which are draw at random, out of which the second ball is red, then we need to find the probability that first ball drawn is also red.
Let us consider that the probability of the first ball drawn is red P(A).
Let us consider that the probability of the second ball drawn is red P(B).
If the second ball is red, then we have two cases where either the first ball is red or black.
Now, by using the formula of probability = (no. of outcomes) / (Total no. of outcomes)
Therefore, P(B) = (first ball is black and second ball is red) + (first ball is red and second ball is red)
                            = $\left( \dfrac{^{7}{{\text{C}}_{1}}}{10}\times \dfrac{^{3}{{\text{C}}_{1}}}{9} \right)+\left( \dfrac{^{3}{{\text{C}}_{1}}}{10}\times \dfrac{^{2}{{\text{C}}_{1}}}{9} \right)$
Here, $\dfrac{^{7}{{\text{C}}_{1}}}{10}$ means, out of 7 black balls 1 is selected from the total number of 10 balls and $\dfrac{^{3}{{\text{C}}_{1}}}{9}$ means out of 3 red balls 1 ball is selected from the remaining number of balls that is 9.
Similarly, it is in the second case, where second ball drawn is red which is out of 3 red balls and the first ball is red which is out of 2 remaining red balls.
P(B) $=\left( \dfrac{7}{10}\times \dfrac{3}{9} \right)+\left( \dfrac{3}{10}\times \dfrac{2}{9} \right)$
        $\begin{align}
  & =\left( \dfrac{21}{90} \right)+\left( \dfrac{6}{90} \right) \\
 & =\dfrac{21+6}{90} \\
 & =\dfrac{27}{90} \\
\end{align}$
Therefore, probability that the second ball is red is $\dfrac{3}{10}$.
Now, we need to use conditional probability to find the probability that the first ball draw is red when the second ball drawn is red.
Therefore, probability that the first is a red ball and the second ball is also red,
P(A/B) = $\dfrac{\text{P}\left( \text{A}\cap \text{B} \right)}{\text{P}\left( \text{B} \right)}$
            = $\dfrac{\dfrac{^{3}{{\text{C}}_{1}}}{10}\times \dfrac{^{2}{{\text{C}}_{1}}}{9}}{\dfrac{3}{10}}$
            = $\dfrac{\dfrac{3}{10}\times \dfrac{2}{9}}{\dfrac{3}{10}}$
            = $\dfrac{\dfrac{6}{90}}{\dfrac{3}{10}}$
            = $\dfrac{6}{90}\times \dfrac{10}{3}$
            = $\dfrac{2}{9}$

Hence, the probability that the first ball drawn is red when the second ball drawn is red is $\dfrac{2}{9}$.

Note: Combination ‘C’ is a selection of items from a collection, such that the order of the selection does not matter. Probability is the possible number of positive outcomes according to the required result.