Question

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?

Answer 1

i think we can find probability for NONE of them will pass.
and subtract it from 1

Answer 2

probability for none of them passing = (1-3/5)(1-3/28)(1-3/10)

Answer 3

So it's 2/5 * 25/28 * 7/10?

Answer 4

1- that

Answer 5

So it's 7/8?

Answer 6

wolfram says 3/4

Answer 7

pls do check if there is some flaw in my logic.. . im feeling slight dyslexic today...

Answer 8

Let's see...your logic is correct, but I don't understand it...

Answer 9

i thought like this :

P(atleast one of them passing) + P(none of them passing) = 1

Answer 10

P(atleast one of them passing) = 1 - P(none of them passing)

Answer 11

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...

Answer 12

But why did you have me subtract one form A, B, and C and then subtract again?

Answer 13

becos, P(A) + P(A') = 1

Answer 14

we are given P(A) <---- p(A passing)
we need P(A') <------ p(A not passing)

Answer 15

P(atleast one of them passing) = 1 - P(none of them passing)
P(none of them passing) = P(A') . P(B') . P(C')

Answer 16

we have used both above equations

Answer 17

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?

Answer 18

thats exactly what we did :)

Answer 19

ALright THank you :)

Answer 20

np.. :)