Question

A bag contains \[3\] black and \[4\] white balls. Two balls are drawn one by one at random without replacement. The probability that the second drawn ball is white
 \[A){\text{ }}\dfrac{4}{{49}}\]
\[B){\text{ }}\dfrac{1}{7}\]
\[C){\text{ }}\dfrac{4}{7}\]
\[D){\text{ }}\dfrac{{12}}{{49}}\]

Answer 1

Hint: In this question we have given a bag from which two balls are drawn randomly without any replacement. First calculate the total number of balls in a bag, then find the probability in two different ways. And then add those two values obtained from two different cases to get the final value.

Complete step by step answer:
It is given that in a bag there are \[3\] black balls and \[4\] white balls . Therefore,
Total number of balls that a bag contains \[ = {\text{ }}3 + 4{\text{ }} = {\text{ }}7\]
There are two possible cases to find the probability that the second drawn ball is white.
In the first case, if the first ball is black ball and the second ball is white ball, then the probability that the second drawn ball is white will be
Probability \[ = {\text{ }}\left( {\dfrac{3}{7}} \right) \times \left( {\dfrac{4}{6}} \right)\] \[ = \] \[\dfrac{2}{7}\]
Because the probability that the black ball is drawn first is \[\dfrac{3}{7}\]. Also it is given that replacement of the balls is not allowed. Therefore, the probability that the white ball is drawn at second is \[\dfrac{4}{6}\]
Now in the second case, if the first ball is white ball and the second ball is black ball, then the probability that the second drawn ball is white will be
Probability \[ = {\text{ }}\left( {\dfrac{4}{7}} \right) \times \left( {\dfrac{3}{6}} \right) = {\text{ }}\dfrac{2}{7}\].
Because the probability that the white ball is drawn first is \[\dfrac{4}{7}\]. Also it is given that replacement of the balls is not allowed. Therefore, the probability that the black ball is drawn at second is \[\dfrac{3}{6}\].
Therefore, the probability that the second drawn ball is white \[ = {\text{ }}\dfrac{2}{7} + \dfrac{2}{7} = {\text{ }}\dfrac{4}{7}\].
Hence, the probability that the second drawn ball is white is \[\dfrac{4}{7}\].

Note:
Probability without replacement means once we draw an item, then we do not replace it back to the sample space before drawing a second item. In other words, an item cannot be drawn more than once.