## 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?

1. ganeshie8

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

2. ganeshie8

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

3. Calcmathlete

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

4. ganeshie8

1- that

5. Calcmathlete

So it's 7/8?

6. ganeshie8

wolfram says 3/4

7. ganeshie8

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

8. Calcmathlete

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

9. ganeshie8

i thought like this : P(atleast one of them passing) + P(none of them passing) = 1

10. ganeshie8

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

11. ganeshie8

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...

12. Calcmathlete

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

13. ganeshie8

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

14. ganeshie8

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

15. ganeshie8

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

16. ganeshie8

we have used both above equations

17. Calcmathlete

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?

18. ganeshie8

thats exactly what we did :)

19. Calcmathlete

ALright THank you :)

20. ganeshie8

np.. :)