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    Mr Lucky plays two games.
    The two games are Game A and Game B.
    Playing Game A and Game B are independent events.
    The probability that Mr Lucky wins both games is 9/25.
    The probability that Mr Lucky wins Game B is four times greater than the probability of him losing Game A.
    Find the probability that Mr Lucky wins only one of the two games he plays.

    So I know that P(winA)×P(winB) = 9/25, and that P(winB)=4P(loseA), but I'm not exactly sure how to progress? Any help is appreciated.
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    Make 2 tree diagrams, one Game A and one Game B,
    remembering that they are independent events.
    Call P(Win A) ‘x’ and the P(Lose A) ‘1-x’.

    From here you know that the probability of him winning game B is 4 times the probability of him losing game A. From this you can write down the probability of him losing game B.

    Form a quadratic that is equal to 9/25 and solve for ‘x’.

    Can you take it from here?
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    (Original post by Y11_Maths)
    Make 2 tree diagrams, one Game A and one Game B,
    remembering that they are independent events.
    Call P(Win A) ‘x’ and the P(Lose A) ‘1-x’.

    From here you know that the probability of him winning game B is 4 times the probability of him losing game A. From this you can write down the probability of him losing game B.

    Form a quadratic that is equal to 9/25 and solve for ‘x’.

    Can you take it from here?
    Got it - thank you
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    (Original post by irrelevant kid)
    Got it - thank you
    Want to share answers?
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    (Original post by Y11_Maths)
    Want to share answers?
    P(lose A) = x
    P(win A) = 1-x
    P(win B) = 4x

    4x × 1-x = 4x-4x^2
    4x-4x^2 = 9/25
    Multiply and rearrange:
    100x^2-100x+9=0
    (10x-1)(10x-9)=0
    x=1/10 or x=1/9
    Used 1/10 as worked.

    Win A, lose B = 54/100
    Lose A, win B = 4/100
    P(win only one) = 58/100 = 29/50

    Going to bed now, so if I've done anything wrong just reply and I'll correct tomorrow.
    Thanks for helping by the way, I appreciate it.
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    (Original post by irrelevant kid)
    P(lose A) = x
    P(win A) = 1-x
    P(win B) = 4x

    4x × 1-x = 4x-4x^2
    4x-4x^2 = 9/25
    Multiply and rearrange:
    100x^2-100x+9=0
    (10x-1)(10x-9)=0
    x=1/10 or x=1/9
    Used 1/10 as worked.

    Win A, lose B = 54/100
    Lose A, win B = 4/100
    P(win only one) = 58/100 = 29/50

    Going to bed now, so if I've done anything wrong just reply and I'll correct tomorrow.
    Thanks for helping by the way, I appreciate it.
    This is correct well done. And I am happy to help anytime
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