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This is a fairly simple problem that can be done by considering all the elements in the sample space and picking out the favorable outcomes but I wanted to know if there was a solution using conditional probability which says for my problem :

P(multiple of 2 on one die AND multiple of 3 on other die)=$P(A).P(B|A)$ where A and B are respective events but I don't know how to proceed. Clearly $P(A)=1/2$ but how do I find the 2nd factor. I am trying to do it this way so that I can generalise the result to a N-face die, as for larger N it is not feasible to count all the favorable outcomes.Thanks!

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  • $\begingroup$ $P(B \mid A) = P(B)$ as $A$ and $B$ are independent $\endgroup$
    – Math Lover
    Commented Feb 28, 2022 at 15:58
  • $\begingroup$ @MathLover Yea but then the probability will be $\frac{1}{6}$ as $P(B)=\frac {1}{3}$ but that isn't the correct answer $\endgroup$
    – Parth Shresth
    Commented Feb 28, 2022 at 16:04
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    $\begingroup$ I just mentioned $P(B \mid A) = P(B)$ but that does not mean $1/6$ or $1/3$ is the answer. You can apply P.I.E as in the below answer. Conditional probability has little meaning here. Btw, you could always write it as $ \small P(B \mid A = 2) \cdot P(A = 2) + P(B \mid A = 3) \cdot P(A = 3) + P(B \mid A = 4) \cdot P(A = 4) + P(B \mid A = 6) \cdot P(A = 6)$ $\endgroup$
    – Math Lover
    Commented Feb 28, 2022 at 16:25

1 Answer 1

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As @Math Lover said in the comments, $\mathbb P(B \mid A) = \mathbb P(B)$ since the events are independent. To make this a full solution, consider the two possible cases for how you might fulfill the event you're looking for. For simplicity, let's color the dice (I'll use "green" and "yellow"):

Case 1: The green die is divisible by 2, and the yellow die is divisible by 3. In this case, we have $\mathbb P(A \cap B) = \mathbb P(A) \cdot \mathbb P(B \mid A) = \mathbb P(A) \cdot \mathbb P(B) = \frac 1 6.$

Case 2: The yellow die is divisible by 2, and the green die is divisible by 3. In this case, the probability is again $\frac 1 6$.

We're concerned with the union of these two cases -- but, they're not disjoint, so we need an inclusion-exclusion principle:

$$\mathbb P(C \cup D) = \mathbb P(C) + \mathbb P(D) - \mathbb P(C \cap D)$$

In other words, we've double counted the situation where cases 1 and 2 simultaneously occur (i.e. that we roll two sixes), so we need to remove that case by subtracting its probability. The final answer should be:

$$ \frac 1 6 + \frac 1 6 - \frac 1 {36} = \fbox{$\frac{11}{36}$}$$

The conditional probability part isn't the difficulty in this problem; decomposing the event into manageable pieces is the obnoxious part.

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