Question

A bag contains 3 white, 3 black and 2 red balls. One by one three balls are drawn without replacing them, then find the probability that the third ball is red.
A.$\dfrac{1}{4}$
B.$\dfrac{3}{4}$
C.$\dfrac{1}{3}$
D.None of these

Answer 1

Hint: To find the probability that the third ball is red, first we have to consider two cases. The first is the probability that the first two draws don’t have a red ball or we can say that zero balls will be red. Then the second has the probability that the first two draws have a red ball or we can say that one ball will be red. After that when we add those probabilities, we will get the answer.

Complete step-by-step answer:
In this question, we have 3 white, 3 black and 3 red balls in a bag and we have to find the probability of drawing the third ball as a red ball.
We can consider three cases here, one is when the first two balls drawn are not red, i.e 0 red balls, or when the first two balls have 1 red ball and the first two balls are 2 red balls. But the case of 2 red balls is not possible as we have only 2 red balls in the bag and we have to find the probability of drawing a third ball as a red ball. So, it is possible that we can draw 0 red or 1 red ball.
Let us consider the first case of 0 red balls. We have 3 white and 3 black, so the possible ways that from the combination of 6, 2 are drawn is $={}^{6}{{C}_{2}}$ . The total possible ways will be $={}^{8}{{C}_{2}}$ . We know that probability is given by favourable outcomes $\div$ total possible outcomes. Here, the probability that the last ball is red $=\dfrac{2}{6}$ .
Then the probability for 0 red will be: $P(0R)=\dfrac{{}^{6}{{C}_{2}}}{{}^{8}{{C}_{2}}}.\dfrac{2}{6}=\dfrac{6\times 5}{8\times 7}.\dfrac{2}{6}=\dfrac{5}{28}$
Now if we choose 1 red ball from 2, the number of ways is $={}^{2}{{C}_{1}}$ . So from 6 different balls choosing 1 ball, the number of ways is $={}^{6}{{C}_{1}}$ . Here, the probability that one ball is red $=\dfrac{1}{6}$ .
Here the probability that 1 ball is red will be: $P(1R)=\dfrac{{}^{2}{{C}_{1}}{}^{6}{{C}_{1}}}{{}^{8}{{C}_{2}}}=\dfrac{2\times 6}{\dfrac{8\times 7}{2\times 1}}.\dfrac{1}{6}=\dfrac{1}{14}$
Now we can obtain the probability that the third ball is red by adding the terms as $P=P(0R)+P(1R)=\dfrac{5}{28}+\dfrac{1}{14}=\dfrac{1}{4}$
Hence, the probability that the third ball is red is $\dfrac{1}{4}$ .

Note: Your concept regarding the probability should be clear. You must note that there is no need to consider the third case here as only 2 red balls are there. You must take utmost care while applying the formula for combination and make sure that no terms are missing. Avoid silly mistakes while simplifying the probabilities. Also, the common mistake that can be committed here is by multiplying the probabilities at the end instead of adding them, which is completely wrong for this question. We add probabilities for mutually exclusive events and multiply them for independent events.