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Shiming, Doreen, and Brian are studying independently for their driver’s license test. The probability Shiming will pass the test is 3/5, the probability Doreen will pass the test is 3/28, and the probability Brian will pass the test is 3/10. What is the probability at least one of these people passes the test?
 one year ago
 one year ago
Shiming, Doreen, and Brian are studying independently for their driver’s license test. The probability Shiming will pass the test is 3/5, the probability Doreen will pass the test is 3/28, and the probability Brian will pass the test is 3/10. What is the probability at least one of these people passes the test?
 one year ago
 one year ago

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ganeshie8Best ResponseYou've already chosen the best response.1
i think we can find probability for NONE of them will pass. and subtract it from 1
 one year ago

ganeshie8Best ResponseYou've already chosen the best response.1
probability for none of them passing = (13/5)(13/28)(13/10)
 one year ago

CalcmathleteBest ResponseYou've already chosen the best response.1
So it's 2/5 * 25/28 * 7/10?
 one year ago

ganeshie8Best ResponseYou've already chosen the best response.1
pls do check if there is some flaw in my logic.. . im feeling slight dyslexic today...
 one year ago

CalcmathleteBest ResponseYou've already chosen the best response.1
Let's see...your logic is correct, but I don't understand it...
 one year ago

ganeshie8Best ResponseYou've already chosen the best response.1
i thought like this : P(atleast one of them passing) + P(none of them passing) = 1
 one year ago

ganeshie8Best ResponseYou've already chosen the best response.1
P(atleast one of them passing) = 1  P(none of them passing)
 one year ago

ganeshie8Best ResponseYou've already chosen the best response.1
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...
 one year ago

CalcmathleteBest ResponseYou've already chosen the best response.1
But why did you have me subtract one form A, B, and C and then subtract again?
 one year ago

ganeshie8Best ResponseYou've already chosen the best response.1
becos, P(A) + P(A') = 1
 one year ago

ganeshie8Best ResponseYou've already chosen the best response.1
we are given P(A) < p(A passing) we need P(A') < p(A not passing)
 one year ago

ganeshie8Best ResponseYou've already chosen the best response.1
P(atleast one of them passing) = 1  P(none of them passing) P(none of them passing) = P(A') . P(B') . P(C')
 one year ago

ganeshie8Best ResponseYou've already chosen the best response.1
we have used both above equations
 one year ago

CalcmathleteBest ResponseYou've already chosen the best response.1
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?
 one year ago

ganeshie8Best ResponseYou've already chosen the best response.1
thats exactly what we did :)
 one year ago

CalcmathleteBest ResponseYou've already chosen the best response.1
ALright THank you :)
 one year ago
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