A bag contains 5 black, 3 white and 2 red balls. Three balls are drawn in succession. What is the probability that the first ball is red, the second ball is black and the third ball is white?

The correct option is 1/24.

Let us find the probability of drawing a red ball first, a black ball second, and a white ball third, in succession without replacement.
First, we calculate the total number of balls in the bag:
Total balls = 5 black + 3 white + 2 red = 10 balls.

Step 1: Probability of drawing the first ball as red
There are 2 red balls out of a total of 10 balls.
The probability of drawing a red ball first is:
P(Red first) = $\frac{2}{10}$ = $\frac{1}{5}$

Step 2: Probability of drawing the second ball as black
Since one red ball has been drawn, the remaining number of balls in the bag is 9.
There are 5 black balls in the bag.
The probability of drawing a black ball second, given that the first ball was red, is:
P(Black second | Red first) = $\frac{5}{9}$

Step 3: Probability of drawing the third ball as white
After drawing the first two balls (one red and one black), the remaining number of balls in the bag is 8.
There are 3 white balls in the bag.
The probability of drawing a white ball third, given the first two draws, is:
P(White third | Red first and Black second) = $\frac{3}{8}$

Step 4: Combined Probability
The probability of all three events occurring in succession is the product of their individual probabilities:
Total Probability = P(Red first) × P(Black second) × P(White third)
Total Probability = $\frac{2}{10}\times \frac{5}{9}\times \frac{3}{8}$
Total Probability = $\frac{1}{5}\times \frac{5}{9}\times \frac{3}{8}$
Total Probability = $\frac{1\times 5\times 3}{5\times 9\times 8}$
By cancelling out the common factor of 5 in the numerator and denominator, we get:
Total Probability = $\frac{3}{9\times 8}$ = $\frac{3}{72}$
Simplifying the fraction by dividing both the numerator and the denominator by 3, we get:
Total Probability = $\frac{1}{24}$