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Calcmathlete
 3 years 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?
Calcmathlete
 3 years 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?

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ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1i think we can find probability for NONE of them will pass. and subtract it from 1

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1probability for none of them passing = (13/5)(13/28)(13/10)

Calcmathlete
 3 years ago
Best ResponseYou've already chosen the best response.1So it's 2/5 * 25/28 * 7/10?

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1pls do check if there is some flaw in my logic.. . im feeling slight dyslexic today...

Calcmathlete
 3 years ago
Best ResponseYou've already chosen the best response.1Let's see...your logic is correct, but I don't understand it...

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1i thought like this : P(atleast one of them passing) + P(none of them passing) = 1

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1P(atleast one of them passing) = 1  P(none of them passing)

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1P(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...

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

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1becos, P(A) + P(A') = 1

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1we are given P(A) < p(A passing) we need P(A') < p(A not passing)

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1P(atleast one of them passing) = 1  P(none of them passing) P(none of them passing) = P(A') . P(B') . P(C')

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1we have used both above equations

Calcmathlete
 3 years ago
Best ResponseYou've already chosen the best response.1Oh. 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?

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1thats exactly what we did :)

Calcmathlete
 3 years ago
Best ResponseYou've already chosen the best response.1ALright THank you :)
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