I'm presented with the following conditional probability problem:
There are 2 bags with different numbers of lemon and grape candies:
- Bag #1: 5 lemon candies, 5 grape candies (10 candies total)
- Bag #2: 4 lemon candies, 8 grape candies (12 candies total)
A person then performs the following events using the two bags in succession (i.e. dependent, yet mutually exclusive):
- $\mathrm{Event \ I}$ - Randomly pick a candy from Bag #1 and place it in Bag #2. (hence, Bag #1 now has 1 less candy and Bag #2 has 1 more)
- $\mathrm{Event \ J}$ - Randomly pick another candy from Bag #1 and place it in Bag #2 (hence, Bag #1 now has 2 less candy, and Bag #2 has 2 more)
- $\mathrm{Event \ K}$ - Randomly pick the final candy from Bag #2.
Calculate $P(A)$, $P(B)$ and $P(A \cap B)$ given that $\mathrm{Event \ A}$ and $\mathrm{Event \ B}$ are described as follows:
- $\mathrm{Event \ A}$ - 2 similar-flavored candies are picked from Bag #1.
- $\mathrm{Event \ B}$ - The candy picked from Bag #2 is grape flavoured.
Here's my attempt at calculating the required probabilities (I haven't simplified the probabilities to keep the denominators similar to that of the total candies in each bag):
- Link to image: Calculating P(A), P(B) and $P(A \cap B)$ w/ a tree diagram
Based on the working given in the image, my answers are as follows:
- $P(A) = P(I \cap J) + P(\bar{I} \cap \bar{J}) = 2(\frac{5}{10} \times \frac{4}{9}) = \frac{4}{9}$
- $P(B) = P(\bar{K}) = P(\bar{K} \cap J \cap I) + P(\bar{K} \cap J \cap \bar{I}) + P(\bar{K} \cap \bar{J} \cap I) + P(\bar{K} \cap \bar{J} \cap \bar{I}) = \frac{9}{14}$
- $P(A \cap B) = P(\bar{K} \cap J \cap I) + P(\bar{K} \cap \bar{J} \cap \bar{I}) = \frac{2}{7}$
Here's the problem: How come $P(A \cap B)$, i.e. $\frac{2}{7}$ is larger than $P(B)$, i.e. $\frac{9}{14}$?
Any corrections or advice will be highly appreciated.
Specifically speaking of
if by this you mean "Can it be that $P(A \cap B) \gt P(A)$ or $P(A \cap B) \gt P(B)$?", the answer is no, by monotonicity of measures