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

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

So my understanding and this explainer leads me to believe, that P(FS)=P(B)=P(F)P(S)=0.18.P(F \cap S) = P(B) = P(F) \cdot P(S) = 0.18. \\ Also, I was quite confident that P(Q)=P(F\B)=P(F)P(B)=0.42. 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)=0.5P(Q)=0.5.

What was my mistake?

Note by Christian Grosser
7 months, 3 weeks ago

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1 vote

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

It should

P(B)=(PS)P(S)=(10.8)0.3=0.5P(B) = (P \cup S) - P(S) = (1 - 0.8) - 0.3 = 0.5 .

Pi Han Goh - 7 months, 3 weeks ago

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P(B)=P(FS)P(B) = P(F \cap S) is not 0.50.5, P(F\S)P(F \backslash S) is. You likely mean P(F\S)=P(FS)P(S)=10.20.3=0.5 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 - 7 months, 3 weeks ago

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