Confusing probabilities with inclusion-exclusion

Hello,

I have a question about this practice problem. The explanation of the solution to the problem given on that page makes intuitive sense to me, but I got to a different solution and can't see where I went wrong.

The problem concerns a selection of students with the probability of each student being enrolled in

French being $P(F)=0.6$
Spanish being $P(S)=0.3$
neither being $P(\overline{F \cup S})=0.2$ .

It asks for the probability of one student being in French but not in Spanish $P(F\backslash S) = P(Q)$ .

So my understanding and this explainer leads me to believe, that $P(F \cap S) = P(B) = P(F) \cdot P(S) = 0.18. \\$ Also, I was quite confident that $P(Q) = P(F \backslash B) = P(F)-P(B)=0.42. \\$ But that is of course different from the intuitively explained solution of $P(Q)=0.5$ .

What was my mistake?

#Combinatorics

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Comments

The probability of a student studying french or spanish is not mutually exclusive, so $P(B) = P(F) \cdot P(S)$ does not hold.

It should

$P(B) = (P \cup S) - P(S) = (1 - 0.8) - 0.3 = 0.5$ .

Pi Han Goh - 6 months, 1 week ago

$P(B) = P(F \cap S)$ is not $0.5$ , $P(F \backslash S)$ is. You likely mean $P(F \backslash S) = P(F \cup S) - P(S) = 1-0.2-0.3=0.5$ . But that clears things up. Thank you.

Christian Grosser - 6 months, 1 week ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$