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The following table represents a distribution of candies in a bag. Each candy has two properties: Its texture and its color.

Distribution of Candies

Question:

If I draw two candies without replacement, what is the probability of first drawing a blue candy and then drawing a hard candy?

Attempt:

Let $B = \text{blue}$ , $H = \text{hard}$, $R = \text{red}$, $S = \text{soft}$

I first considered multiplying the marginal probability of $B$ with the marginal probability of $H$ (with one less in the denominator to take into account non-replacement):

$$\frac{11}{28} \cdot \frac{13}{27} = \frac{143}{756}$$

It then occurred to me that the first candy drawn could be both blue and hard which would effect $P(H)$ during the second draw as illustrated in the probability tree.

+----------------+-----------------------+
| If first candy | P(H) on second draw = |
|  is blue and   |                       |
+----------------+-----------------------+
|    Hard        |         12/27         |
+----------------+-----------------------+
|    Soft        |         13/27         |
+----------------+-----------------------+

Probability Tree of Both Draws

I assume the correct answer would be to add both cases in the tree:

$$\frac{60}{756} + \frac{78}{756} = \frac{138}{756}$$

3 Answers 3

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Yes, that looks fine.   It is the Law of Total Probability at work. $$\mathsf P(B_1\cap H_2) ~=~ \mathsf P(B_1\cap H_1)~P(H_2\mid B_1\cap H_1)+\mathsf P(B_1\cap S_1)~\mathsf P(H_2\mid B_1\cap S_1) ~=~ \frac{5\cdot 12+6\cdot 13}{28\cdot 27}$$

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Yes, this is exactly right.....

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Yes its perfect.

We have two cases and sum of them is answer. As you found.