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Calcmathlete

  • 2 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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  1. ganeshie8
    • 2 years ago
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    i think we can find probability for NONE of them will pass. and subtract it from 1

  2. ganeshie8
    • 2 years ago
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    probability for none of them passing = (1-3/5)(1-3/28)(1-3/10)

  3. Calcmathlete
    • 2 years ago
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    So it's 2/5 * 25/28 * 7/10?

  4. ganeshie8
    • 2 years ago
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    1- that

  5. Calcmathlete
    • 2 years ago
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    So it's 7/8?

  6. ganeshie8
    • 2 years ago
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    wolfram says 3/4

  7. ganeshie8
    • 2 years ago
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    pls do check if there is some flaw in my logic.. . im feeling slight dyslexic today...

  8. Calcmathlete
    • 2 years ago
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    Let's see...your logic is correct, but I don't understand it...

  9. ganeshie8
    • 2 years ago
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    i thought like this : P(atleast one of them passing) + P(none of them passing) = 1

  10. ganeshie8
    • 2 years ago
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    P(atleast one of them passing) = 1 - P(none of them passing)

  11. ganeshie8
    • 2 years ago
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    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
    • 2 years ago
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    But why did you have me subtract one form A, B, and C and then subtract again?

  13. ganeshie8
    • 2 years ago
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    becos, P(A) + P(A') = 1

  14. ganeshie8
    • 2 years ago
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    we are given P(A) <---- p(A passing) we need P(A') <------ p(A not passing)

  15. ganeshie8
    • 2 years ago
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    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
    • 2 years ago
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    we have used both above equations

  17. Calcmathlete
    • 2 years ago
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    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
    • 2 years ago
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    thats exactly what we did :)

  19. Calcmathlete
    • 2 years ago
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    ALright THank you :)

  20. ganeshie8
    • 2 years ago
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    np.. :)

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