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probability heeeeeelp

hi, p(A) = 0.3 and P(AuB) = 0.7 no diagram given. Find P(A'nB). answer is 0.4. My question is, so if we were to draw a venn diagram, our outside bit is therefore 0.3, then the whole of a, so a plus intersection ( x ) is 0.3. let's say that just B ( not including AnB) is called y. Then we know that P(AuB) = 0.7 = P(A) + P(B) - P(AnB) = 0.3 + x+ y -x. Therefore 0.7 = 0.3 + y so 0.4 = y , but that's not possible if we have an intersection? Then there wouldn't be an intersection??? I'm going mad, it's question 3b on this sheet https://www.mathsgenie.co.uk/resources/a-stats-probability.pdf and here are the answers https://www.mathsgenie.co.uk/resources/a-stats-probabilityans.pdf
WAIT OH MY GOD I GOT IT, ESSENTIALLY IT'S THAT CRESCENT A IS 0.1 AND THEN INTERSECT IS 0.2 THEY ADDED THAT THEY WERE INDEPENDENT SO I SIMULTANEOUS EQUATIONED IT aHAHAHAAH

but id still like some help because apparently the method is much simpler than my one page of working whoops
ok no 3 c and d are literally on meth, I get 3c now after extreme amounts of work, but wtf is up with d, (A'nB') is the outside bit, it's not everything except for a intercept b, that would be (AnB)'. am I going mad or is this question wrong???
Original post by the bear
vd.png

the green bit + the purple bit = o.3

the green bit + the purple bit + the blue bit = 0.7

so the blue bit = 0.4

we want to find "not in A, and in B" or if you like "people who do not like Apples but do like Bananas"

in other words the blue bit.

p(A' B) = 0.4

thanks so so much, I get that now, I'm just completely lost on d. P(A'uB') is apparently 1- P(AnB) so 1-6/35???? but that's wrong, it should be written P(AnB)' in the question
Reply 4
vd.png

the green bit + the purple bit = 0.3

the green bit + the purple bit + the blue bit = 0.7

so the blue bit = 0.4

we want to find "not in A, and in B" or if you like "people who do not like Apples but do like Bananas"

in other words the blue bit.

p(A' B) = 0.4

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