<p>Box I: 3 red, 6 black; Box II: 5 red, 5 black. A box is selected at random and 2 balls are drawn. Both are red. Which of the following about \(P(\text{Box I} \mid \text{both red})\) are TRUE?</p>
P(Box I | both red) = 1/7
P(Box I | both red) < 1/2
P(both red | Box I) = 1/12
P(both red | Box II) = 2/9
Step-by-Step Solution
Key Concept: Apply Bayes' theorem. P(both red|I) = C(3,2)/C(9,2) = 3/36 = 1/12; P(both red|II) = C(5,2)/C(10,2) = 10/45 = 2/9.
Step 1: Calculate the probability of drawing two red balls given Box I was selected.
Box I contains 3 red balls and 6 black balls, for a total of 9 balls. The probability of drawing 2 red balls from Box I, denoted as $P(\text{both red | Box I})$, is calculated using combinations.
$$P(\text{both red | Box I}) = \frac{\binom{3}{2}}{\binom{9}{2}}$$
$$P(\text{both red | Box I}) = \frac{\frac{3 \times 2}{2 \times 1}}{\frac{9 \times 8}{2 \times 1}} = \frac{3}{36} = \frac{1}{12}$$
This confirms that Option 3: $P(\text{both red | Box I}) = 1/12$ is TRUE.
Step 2: Calculate the probability of drawing two red balls given Box II was selected.
Box II contains 5 red balls and 5 black balls, for a total of 10 balls. The probability of drawing 2 red balls from Box II, denoted as $P(\text{both red | Box II})$, is calculated using combinations.
$$P(\text{both red | Box II}) = \frac{\binom{5}{2}}{\binom{10}{2}}$$
$$P(\text{both red | Box II}) = \frac{\frac{5 \times 4}{2 \times 1}}{\frac{10 \times 9}{2 \times 1}} = \frac{10}{45} = \frac{2}{9}$$
This confirms that Option 4: $P(\text{both red | Box II}) = 2/9$ is TRUE.
Step 3: Calculate the overall probability of drawing two red balls.
Since one of the boxes is selected at random, the probability of selecting Box I is $P(\text{Box I}) = 1/2$, and the probability of selecting Box II is $P(\text{Box II}) = 1/2$. We use the Law of Total Probability to find the overall probability of drawing two red balls, $P(\text{both red})$.
$$P(\text{both red}) = P(\text{both red | Box I})P(\text{Box I}) + P(\text{both red | Box II})P(\text{Box II})$$
$$P(\text{both red}) = \left(\frac{1}{12}\right)\left(\frac{1}{2}\right) + \left(\frac{2}{9}\right)\left(\frac{1}{2}\right)$$
$$P(\text{both red}) = \frac{1}{2}\left(\frac{1}{12} + \frac{2}{9}\right) = \frac{1}{2}\left(\frac{3}{36} + \frac{8}{36}\right) = \frac{1}{2}\left(\frac{11}{36}\right) = \frac{11}{72}$$
Step 4: Calculate the probability that Box I was selected given that two red balls were drawn.
We use Bayes' Theorem to find $P(\text{Box I | both red})$.
$$P(\text{Box I | both red}) = \frac{P(\text{both red | Box I})P(\text{Box I})}{P(\text{both red})}$$
$$P(\text{Box I | both red}) = \frac{\left(\frac{1}{12}\right)\left(\frac{1}{2}\right)}{\frac{11}{72}}$$
$$P(\text{Box I | both red}) = \frac{\frac{1}{24}}{\frac{11}{72}} = \frac{1}{24} \times \frac{72}{11} = \frac{3}{11}$$
Step 5: Evaluate the given options.
Based on our calculations:
* $P(\text{Box I | both red}) = 3/11$.
* $P(\text{both red | Box I}) = 1/12$.
* $P(\text{both red | Box II}) = 2/9$.
Let's check each option:
* Option 1: $P(\text{Box I | both red}) = 1/7$. This is FALSE, as $3/11 \neq 1/7$.
* Option 2: $P(\text{Box I | both red}) < 1/2$. This is TRUE, as $3/11 \approx 0.27 < 0.5$.
* Option 3: $P(\text{both red | Box I}) = 1/12$. This is TRUE.
* Option 4: $P(\text{both red | Box II}) = 2/9$. This is TRUE.
The options that are TRUE based on the calculations are Option 2, Option 3, and Option 4.
The final answer is $\boxed{\text{ABC}}$
Correct Answer: ABC