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From this question: The probability that a student passes Math 31 is $\frac23$, and the probability that the student passes English is $\frac49$. If the probability of passing at least one course is $\frac45$, what is the probability that he passes both exam?

Then my teacher used the formula of addition rule by simply transposing the terms that become like this:

$$P(M\cap E) = P(M) + P(E) - P(M\cup E)$$

Note: Sorry for poor construction, i am writing this through mobile and having little access on some other tools.

My questions are:

  1. why did my teacher used that formula instead of multiplication rule wherein we are looking for the probability of BOTH passing the exam? And

  2. how to distinguish from the question what exact formula to use?

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You can't use the multiplication rule directly because you are not told that the two events are independent events. If $A$ and $B$ are independent, then we have $P(A \cap B)=P(A)P(B)$, in general, this is not true.

For this question, you are given $P(M \cup E)$, $P(M)$, and $P(E)$ and you want to find $P(M \cap E)$. The formula links all these quantities together, hence you should use it.

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  • $\begingroup$ How come passing both exam is not independent? $\endgroup$
    – log
    Commented Oct 5, 2019 at 5:33
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    $\begingroup$ You are not given the information so you can't be sure until you verify it. In practice, perhaps the ability to do well in maths maths and the ability to do well in English can be positively correlated or negatively correlated isn't it? Or they might be have any relationship at all. $\endgroup$
    – Siong Thye Goh
    Commented Oct 5, 2019 at 5:39
  • $\begingroup$ $\frac 45 = \frac{36}{45} = P(M \cup E) = P(M) + P(E) - P(M\cap E) = \frac{30}{45} + \frac{20}{45} - P(M\cap E).$ $\endgroup$
    – BruceET
    Commented Oct 5, 2019 at 7:45

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