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John has a 45% chance of passing math. A 30% chance of passing both math and statistics. A 70% chance of passing either math or statistics or both. John has been informed that he has passed statistics, what is the probability that he will pass math?

a. 0.58
b. 0.88
c. 0.55
d. 0.70
e. 0.15

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4 Answers 4

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Without formulas:

We know these:

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But these gives:

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So we have cleared that:

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We know that stat is true:

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So the only formula needed is the one for independent events:

$P_{+Math}=\frac{30}{30+25}=\frac{30}{55}=0.\overline{54}$

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Let $S$ be the event John passes Statistics and $M$ be the event John passes math.

You have $P(S \cap M)=.3, P(M)=.45, P(S \cup M)=.7$.

Apply the Conditional probability formula $$P(A | B)=\frac{P(A \cap B)}{P(B)}$$ and Probability for union of 2 events (with some rearrangement) $$P(A \cup B)=P(A)+P(B) - P(A \cap B)$$

to find that $P(M|S)$ is approximately .55

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    $\begingroup$ It is okay. I get it. $\endgroup$ Jan 18, 2018 at 20:24
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    $\begingroup$ If it helped, please accept/upvote. $\endgroup$
    – sma
    Jan 18, 2018 at 20:26
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Skim's answer is precise and efficient. If you are more visually oriented, you can do the same logic with the help of a venn diagram.
Venn diagram

If we already know that John passed stats, that means that he is in the blue set, so we are now only dealing with that 55% of the total. Think of it as 55 out of 100 total students. Then 30 out of those 55 also passed math, which makes 30/55 = .55 or 55% probability.

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Take $A$ as chance of passing math and $B$ as that of statistics. Therefore we have: $$p(A)=0.45\\p(B)=0.3\\p(A\cup B)=0.7$$ Now the $p(A|B)$ has been requested which is: $$p(A|B)=\frac{p(A\cap B)}{p(B)}=\frac{p(A)+p(B)-p(A\cup B)}{p(B)}=\frac{1}{6}$$

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