Question 254938: John and Jane are married. The probability than John watches a certain television show is 0.4. The probability that Jane watches the show is 0.5. The probability that John watches the show, given that Jane does, is 0.7.
a) find the probability that both John and Jane watch the show.
b) find the probability that Jane watches the show, given that John does.
c) do John and Jane watch the show independently of each other? Justify your answer.
I tried the set with
{ WW, WD, DW, DD} with W being 'watched' and D standing for 'did not watch.'
It came out that a was 1/2 x 1/2 = .25
I also tried multiplying the probabilities together and got 0.20/25 = .8
That was for a. B would be .7/5 or .7 x .5 = .35. Is that even correct?
I say they could watch it independently of each other if there were other factors, but since they aren't part of the equation, I would say no.
Could you sort of give me a nudge?
Answer by edjones(8007)   (Show Source):
You can put this solution on YOUR website! j=probability john watches
m= probability jane watches
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Check the Venn diagram below
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Lets take a weekly show for 10 weeks. j watches 4 shows and m watches 5 in that period. The probability that John watches the show, given that Jane does, is 0.7 This means that while m is watching one of her five shows j is also watching .7 of the time so he is watching 5*.7=3.5 shows. That leaves him only 0.5 shows to watch on his own.
Now we divide the above numbers by 10 and do the Venn diagram. Put .05 in the red circle, .15 in the green circle and .35 in the intersection between the circles and .45 outside the circles for when nobody is watching the show so that the sum equals 1
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a) .35 the intersection.
b) .35/.4=.875
c) Yes. .05 for j and 1.5 for m
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E-mail me any more questions.
This is a tough one. Let me know if it is correct or not.
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Ed
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