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The question was the following:

There are $28$ students in a class, $15$ study chemistry, $18$ study physics and $2$ study neither chemistry nor physics.

Calculate the probability that a student chosen at random studies both chemistry and physics.

My approach was as follows:

We pick a random person, the chance that this person studies physics is $\frac{18}{28}$ (because $18$ people study physics out a total of $28$). Now given that he studies physics, what is the chance that he also studies chemistry? I would say $P(\text{chem}|\text{phy}) = \frac{15}{26}$, because we already now that the person studies physics we can exclude the two persons who study neither given $26$ in total, of which $15$ study physics. Concluding that: $P(\text{phy and chem}) = P(\text{phy})P(\text{chem}|\text{phy}) = \frac{18}{28}\frac{15}{26} = 0.37$.

However, the solutions provides elsewjere stated the following approach:

$P(\text{phy and chem}) = P(\text{phy}) + P(\text{chem}) - P(\text{phy}\cup \text{chem})$. Where $P(\text{phy}) = \frac{18}{28}$, $P(\text{chem}) = \frac{15}{28}$, and $P(\text{phy}\cup \text{chem}) = 1 - P(\overline{\text{phy}\cup \text{chem}}) = 1 - P(\overline{\text{phy}} \cap \overline{\text{chem}}) = 1 - \frac{2}{28} = \frac{26}{28}$. Concluding that: $P(\text{phy and chem}) = \frac{18}{28}+\frac{15}{28} - \frac{26}{28} = \frac{1}{4} = 0.25$.

The two answers differ. I think I made a mistake in calculating the conditional probability but I can't identify where. Please comment on this.

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I think that N. F. Taussig's answer is spot on, including his demonstration that the OP's analysis is inaccurate. My answer explores the OP's mis-step more deeply.

Let A denote the event that the student studies chemistry.
Let B denote the event that the student studies physics.
Let C denote the event that the student studies chemistry or physics or both.

Then 15/26 = p(A|C), not p(A|B). Meta-cheating, the OP assumed (correctly) that the problem provides sufficient info for an answer, and assumed (incorrectly) that the solution must involve bayes theorem :
p(A|B) x p(B) = p(AB).

Since there didn't seem to be any (other) way of directly computing p(A|B), the OP assumed (incorrectly) that 15/26 = p(A|B).

For someone new to conditional probability (i.e. bayes theorem), this is an unfair problem. Conditional probability problems are tricky and require a step by step development of the student's intuition.

BTW... I was impressed with the OP's metacheating, so I upvoted.

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Let $P$ be the event that a student studies physics; let $C$ be the event that a student studies chemistry. The number who study physics or chemistry is $$|P \cup C| = |P| + |C| - |P \cap C|$$ We know that of the $28$ students, $28 - 2 = 26$ study physics or chemistry. We also know that $18$ study physics and $15$ study chemistry. Hence, $|P \cup C| = 26$, $|P| = 18$, and $|C| = 15$. Substituting these values into the above formula enables us to solve for the number who study both subjects. \begin{align*} 26 & = 18 + 15 - |P \cap C|\\ |P \cap C| & = 18 + 15 - 26\\ |P \cap C| & = 7 \end{align*} Therefore, \begin{align*} \Pr(P) & = \frac{18}{28}\\ \Pr(P \cap C) & = \frac{7}{28} \end{align*} Since $$\Pr(P \cap C) = \Pr(P)\Pr(C \mid P)$$ we may conclude that $$\Pr(C \mid P) = \frac{\Pr(P \cap C)}{\Pr(P)} = \frac{\frac{7}{28}}{\frac{18}{28}} = \frac{7}{18}$$ so your claim that $\Pr(C \mid P) = \frac{15}{26}$ was false as just $7$ of the $18$ students who study physics also study chemistry.

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