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i think we can find probability for NONE of them will pass. and subtract it from 1
probability for none of them passing = (1-3/5)(1-3/28)(1-3/10)
So it's 2/5 * 25/28 * 7/10?
So it's 7/8?
wolfram says 3/4
pls do check if there is some flaw in my logic.. . im feeling slight dyslexic today...
Let's see...your logic is correct, but I don't understand it...
i thought like this : P(atleast one of them passing) + P(none of them passing) = 1
P(atleast one of them passing) = 1 - P(none of them passing)
P(none of them passing) = P(A') . P(B') . P(C') note that these are mutually exclusive evens. one guy passing doesnt depend on other guys passing...
But why did you have me subtract one form A, B, and C and then subtract again?
becos, P(A) + P(A') = 1
we are given P(A) <---- p(A passing) we need P(A') <------ p(A not passing)
P(atleast one of them passing) = 1 - P(none of them passing) P(none of them passing) = P(A') . P(B') . P(C')
we have used both above equations
Oh. ok...now I get it...first you subtract the pass rate from 1 to get the non pass rate individually. Then multiply them to get the combined no pass rate, then subtract from 1 again to get the combined pass rate?
thats exactly what we did :)
ALright THank you :)