Question

A bag contains 2 white, 3 green, and 5 red balls. If three balls are drawn one after the other without replacement, then the probability that the last ball drawn was red is:

• A. \(\frac{2}{3}\)
• B. \(\frac{3}{4}\)
• C. \(\frac{5}{9}\)
• D. \(\frac{1}{2}\)

Hint:

For drawing without replacement, reduce the total number of items after each draw.

Answer 1

The Correct Option is D

Step 1: Determine the total number of balls. Total number of balls = 2 (white) + 3 (green) + 5 (red) = 10 balls. Step 2: Calculate the probability that the third ball is red.
We can consider the possible scenarios for the first two balls and the third ball: Scenario 1: Red ball on the third draw. We can calculate the probability directly as follows: Let R be the event that the third ball is red. We can consider the position of the red ball as fixed in the third position.
The probability that the third ball is red is the same as the probability that the first ball is red. P(R) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{5}{10} = \frac{1}{2} Alternatively, we can compute it as follows: Total ways of drawing 3 balls = 10 \times 9 \times 8 Ways to draw red on the third draw: \begin{itemize} \item Case 1: WW R: 2 \times 1 \times 5 = 10 \item Case 2: WG R: 2 \times 3 \times 5 = 30 \item Case 3: WR R: 2 \times 5 \times 4 = 40 \item Case 4: GW R: 3 \times 2 \times 5 = 30 \item Case 5: GG R: 3 \times 2 \times 5 = 30 \item Case 6: GR R: 3 \times 5 \times 4 = 60 \item Case 7: RW R: 5 \times 2 \times 4 = 40 \item Case 8: RG R: 5 \times 3 \times 4 = 60 \item Case 9: RR R: 5 \times 4 \times 3 = 60 \end{itemize} Total ways = 10 + 30 + 40 + 30 + 30 + 60 + 40 + 60 + 60 = 360 Total ways of drawing 3 balls = 10 \times 9 \times 8 = 720 Probability = \frac{360}{720} = \frac{1}{2} Therefore, the probability that the last ball drawn was red is \frac{1}{2}.