Question

A bag contains $5$ white and $3$ black balls, and \[4\] are successively drawn out and not replaced; what is the chance that they are alternately of a different color?

Answer 1

Hint:
Probability is defined as the ratio of the number of favorable outcomes by total possible outcomes. For the chance of getting alternate colors, we need to make two cases, one where the white ball is drawn at first and the then second case where the black ball is drawn at first. Now find the probability of these two cases separately and then add them to get an answer.

Complete step by step solution:
Here in this problem, we are given a bag filled with $5$ white and $3$ black balls. Then four balls are successively drawn out without replacing. Now with this information, we need to find out the probability of getting four balls alternately of different colors.
Before starting with the solution we must understand the concept of probability. Probability is a measure of the likelihood of an event to occur. Many events cannot be predicted with total certainty. We can predict only the chance of an event to occur i.e. how likely they are to happen, using it. Probability can range from $0$ to $1$ , where $0$ means the event to be an impossible one and $1$ indicates a certain event.
$ \Rightarrow {\text{Probability}} = \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total possible outcomes}}}}$
 In the given bag, there are total $8$ balls and when we need to arrange different color balls alternately, there are two cases possible.
Case $1$ : When the first ball is drawn out is white. In this case the first and the third ball drawn out is white and the second & fourth ball is black.
Case $2$ : When the first ball is drawn out is black. In this case first & the third ball drawn out is black and the second & fourth ball is white.
As we know that after each drawn the number of possible color balls and total balls will reduce by one.
Solving Case $1$ :
$ \Rightarrow $ Probability of getting alternating colors with the first ball of white color$ = \dfrac{5}{8} \times \dfrac{3}{7} \times \dfrac{4}{6} \times \dfrac{2}{5} = \dfrac{1}{{7 \times 2}} = \dfrac{1}{{14}}$
Similarly for case $2$ :
$ \Rightarrow $ Probability of getting alternating colors with the first ball of black color$ = \dfrac{3}{8} \times \dfrac{5}{7} \times \dfrac{2}{6} \times \dfrac{4}{5} = \dfrac{1}{{7 \times 2}} = \dfrac{1}{{14}}$

Therefore the total probability of both the cases will be$ = \dfrac{1}{{14}} \times \dfrac{1}{{14}} = \dfrac{1}{7}$

Note:
In probability, understanding the question and the situation is always important. Notice here we added and multiplied probabilities. The addition of probability is done when the events are independent but the addition is done to find the total probability of an event. Multiplication of probability is done when the events are occurring simultaneously and together.