Question

A bowl contains eight chips. Three of the chips are red and the remaining five are blue. Two chips are to be drawn successively, at random and without replacement. What is the probability that the first draw results in a red chip and that the second draw results in a blue chip?

• A. 0.27
• B. 0.71
• C. 0.38
• D. 0.57

Hint:

For sequential draws without replacement:
\[ \text{Probability} = \frac{\text{Wanted } 1}{\text{Total } 1} \times \frac{\text{Wanted } 2}{\text{Total } 1 - 1} \]
Ensure you reduce the total count in the denominator for the second fraction.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem involves dependent events, where drawing chips without replacement changes the total number of items and the composition of the remaining pool for the second draw.

Step 2: Key Formula or Approach:
The probability of the joint occurrence of two dependent events, \(R_1\) (red on the first draw) and \(B_2\) (blue on the second draw), is:
\[ P(R_1 \cap B_2) = P(R_1) \times P(B_2 \mid R_1) \]
where \(P(B_2 \mid R_1)\) is the conditional probability of drawing a blue chip second, given that a red chip was drawn first.

Step 3: Detailed Explanation:
We are given:
- Total number of chips in the bowl = 8
- Number of red chips = 3
- Number of blue chips = 5
We want to find the probability of drawing a red chip first, followed by a blue chip, without replacement.
1. First, calculate the probability of drawing a red chip on the first draw, \(P(R_1)\):
Since there are 3 red chips out of 8 total chips:
\[ P(R_1) = \frac{3}{8} \]
2. Next, calculate the conditional probability of drawing a blue chip on the second draw, \(P(B_2 \mid R_1)\):
Since we are drawing without replacement, 1 red chip is removed.
The composition of the bowl for the second draw is now:
- Total chips remaining = \(8 - 1 = 7\)
- Red chips remaining = \(3 - 1 = 2\)
- Blue chips remaining = 5 (unchanged)
Therefore, the probability of drawing a blue chip from this remaining pool is:
\[ P(B_2 \mid R_1) = \frac{5}{7} \]
3. Now, multiply these two probabilities to find the joint probability:
\[ P(R_1 \cap B_2) = P(R_1) \times P(B_2 \mid R_1) \]
\[ P(R_1 \cap B_2) = \frac{3}{8} \times \frac{5}{7} = \frac{15}{56} \]
Convert this fraction to a decimal value:
\[ \frac{15}{56} \approx 0.267857 \]
Rounding to two decimal places:
\[ P(R_1 \cap B_2) \approx 0.27 \]
This matches the first option.

Step 4: Final Answer:
Therefore, the correct option is (A).