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?
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\textbf{Probability of Sequential Events.} When calculating the probability of a sequence of dependent events, remember to adjust the total number of outcomes and the number of favorable outcomes after each event occurs.
Total number of balls in the bag \( = 5 \text{ (black)} + 3 \text{ (white)} + 2 \text{ (red)} = 10 \) balls.
We are drawing three balls in succession. We need to find the probability of the event where the first ball is red, the second ball is black, and the third ball is white.
Probability of the first ball being red:
Number of red balls \( = 2 \)
Total number of balls \( = 10 \)
\( P(\text{1st ball is red}) = \frac{2}{10} \)
After drawing one red ball, there are now 9 balls left in the bag: 5 black, 3 white, and 1 red.
Probability of the second ball being black (given the first was red):
Number of black balls \( = 5 \)
Total number of remaining balls \( = 9 \)
\( P(\text{2nd ball is black} | \text{1st was red}) = \frac{5}{9} \)
After drawing one red and one black ball, there are now 8 balls left in the bag: 4 black, 3 white, and 1 red.
Probability of the third ball being white (given the first was red and the second was black):
Number of white balls \( = 3 \)
Total number of remaining balls \( = 8 \)
\( P(\text{3rd ball is white} | \text{1st was red, 2nd was black}) = \frac{3}{8} \)
The probability of all three events occurring in this sequence is the product of their individual probabilities:
$$ P(\text{1st red, 2nd black, 3rd white}) = P(\text{1st red}) \times P(\text{2nd black} | \text{1st red}) \times P(\text{3rd white} | \text{1st red, 2nd black}) $$
$$ P = \frac{2}{10} \times \frac{5}{9} \times \frac{3}{8} $$
$$ P = \frac{1}{5} \times \frac{5}{9} \times \frac{3}{8} $$
$$ P = \frac{1 \times 5 \times 3}{5 \times 9 \times 8} $$
$$ P = \frac{15}{360} $$
Simplify the fraction by dividing both numerator and denominator by their greatest common divisor, which is 15:
$$ P = \frac{15 \div 15}{360 \div 15} = \frac{1}{24} $$
Therefore, the probability that the first ball is red, the second ball is black, and the third ball is white is \( \frac{1}{24} \).
