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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?

1- that

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)

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)

we have used both above equations

thats exactly what we did :)

ALright THank you :)

np.. :)