SOLUTION: I'm having trouble in my stats class. Could anyone help me with the following question? The probability that a student is taking an English course is 0.60, that a student is taking

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Question 799062: I'm having trouble in my stats class. Could anyone help me with the following question? The probability that a student is taking an English course is 0.60, that a student is taking a chemistry class is 0.30, and that a student is taking both English and chemistry is 0.12.
1. find the probability that a student selected at random is taking either English or chemistry
2. find the probability that a student selected at random is not taking English.
3. find the probability that a student is taking English, given that the student is taking chemistry.
4. are the events "taking English" and 'taking chemistry" independent events? explain why?
Found 2 solutions by Finavon, solver91311:
Answer by Finavon(81) (Show Source):
You can put this solution on YOUR website!
p(E) = 0.6
p(C) = 0.3
p(E&C) = 0.12
a) p(EorC) = 0.6+0.3 = 0.9
b) p(notE) = 1-0.6 = 0.4
c) p(C)*p(E') = p(E&C)
p{E') =
d) not independent.
p(E') < p(E) and p(E&C) < p(E)*p(C) = 0.18

Answer by solver91311(24713) (Show Source):
You can put this solution on YOUR website!

Use a Venn diagram. Universe is all students. Two circles, one for English, one for Chemistry. Overlap is worth 12%, English but NOT Chem is then 60% - 12% = 48%. Chem but NOT English is 30% - 12 % = 18%. Either English or Chem (or both) is 48 + 12 + 18 = 78%. Prob Eng given Chem is 12%...if Eng given Chem then must be both. Independent. The fact that a student is taking one or the other tells you nothing about whether or not they are taking the other. Contrast this with a situation where English was a pre-requisite for taking Chem -- then the probability of taking Chem would be dependent on taking or having taken English.

John

Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it