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Math Help - Urgent Probability and statistics Help answer cheak

  1. #1
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    Exclamation Urgent Probability and statistics Help answer cheak

    Two important members of a cricket team are injured, and each has probability 1/3 of recovering before the match. The recoveries of the two players are independent of each other. If both are able to play then the team has probability 3/4 of winning the match, if only one of them plays then the probability of winning is 1/2 and if neither play the probability of winning is 1/16. What is the probability that the match is won?

    The following is what i have done:

    This is total probablity and we need to find the probabilty that the match is won.

    So A= event one players plays
    B= event both players play
    C= event nither plauers play

    Let D be the event that they win the match.

    P(D)= P(D l A) * P(A) + P(D l B) * P(B) + P(D l C) * P(C)

    P(D l A) = 1/2 P(A)= 1/3
    P(D l B) = 3/4 P(B)= 1/3 * 1/3 = 1/9
    P(D l C) = 1/16 P(C) = (1-1/3)^2 = 4/9

    so i got my answers as 1/2 * 1/3 + 3/4 * 2/3 + 1/16 * 4/9 = 5/18

    Can Someone Please Cheak If My Answer Is Correct As I Am Not Really Sure.

    Thanks
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  2. #2
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    Quote Originally Posted by maths_2468 View Post
    Two important members of a cricket team are injured, and each has probability 1/3 of recovering before the match. The recoveries of the two players are independent of each other. If both are able to play then the team has probability 3/4 of winning the match, if only one of them plays then the probability of winning is 1/2 and if neither play the probability of winning is 1/16. What is the probability that the match is won?

    The following is what i have done:

    This is total probablity and we need to find the probabilty that the match is won.

    So A= event one players plays
    B= event both players play
    C= event nither plauers play

    Let D be the event that they win the match.

    P(D)= P(D l A) * P(A) + P(D l B) * P(B) + P(D l C) * P(C)

    P(D l A) = 1/2 P(A)= 1/3
    P(D l B) = 3/4 P(B)= 1/3 * 1/3 = 1/9
    P(D l C) = 1/16 P(C) = (1-1/3)^2 = 4/9

    so i got my answers as 1/2 * 1/3 + 3/4 * 2/3 + 1/16 * 4/9 = 5/18
    First, although this is not your error, these are all horridly written! To the point that I had difficulty understanding it. For example:
    P(D l A) = 1/2 P(A)= 1/3 is completely wrong! What it means is that P(D|A)= 1/2 P(A) and that (1/2)P(A) is equal to 1/3- which made me think you intended P(A) to be 2/3! That's not true nor is it what you meant.
    P(D|A)= 1/2 P(A) and P(A)= 1//3 is what you meant- do NOT connect things with "=" when they are not equal!

    However, your real problem is that P(A) is NOT 1/3. That is the probability that a specific one of the players recovers. A is the event that one or the other recovers.
    Call the two players "x" and "y". Let X be the event that player x recovers, Y the event that player y recovers. Then P(A)= P(X or Y)= P(X)+ P(Y)- P(X and Y). You are told that P(X)= P(Y)= 1/3 and the two recoveries are independent so P(A)= P(X or Y)= 1/3+ 1/3- 1/9= 2/3- 1/9= 5/9. The probabilty the one or the othe but NOT both, which you are counting separately is 5/9- 1/9= 4/9.

    You could also argue that since the P(X)= P(Y)= 1/3, P(not X)= P(not Y), the probabilities that each does NOT recover, is 2/3. Then the probability that x recovers but not y is (1/3)(2/3)= 2/9 and the probability that y recovers but not x is (2/3)(1/3)= 2/9 and so the probability that one but not the other recovers is 2/9+ 2/9= 4/9.

    Can Someone Please Cheak If My Answer Is Correct As I Am Not Really Sure.

    Thanks
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    First, although this is not your error, these are all horridly written! To the point that I had difficulty understanding it. For example:
    P(D l A) = 1/2 P(A)= 1/3 is completely wrong! What it means is that P(D|A)= 1/2 P(A) and that (1/2)P(A) is equal to 1/3- which made me think you intended P(A) to be 2/3! That's not true nor is it what you meant.
    P(D|A)= 1/2 P(A) and P(A)= 1//3 is what you meant- do NOT connect things with "=" when they are not equal!

    However, your real problem is that P(A) is NOT 1/3. That is the probability that a specific one of the players recovers. A is the event that one or the other recovers.
    Call the two players "x" and "y". Let X be the event that player x recovers, Y the event that player y recovers. Then P(A)= P(X or Y)= P(X)+ P(Y)- P(X and Y). You are told that P(X)= P(Y)= 1/3 and the two recoveries are independent so P(A)= P(X or Y)= 1/3+ 1/3- 1/9= 2/3- 1/9= 5/9. The probabilty the one or the othe but NOT both, which you are counting separately is 5/9- 1/9= 4/9.

    You could also argue that since the P(X)= P(Y)= 1/3, P(not X)= P(not Y), the probabilities that each does NOT recover, is 2/3. Then the probability that x recovers but not y is (1/3)(2/3)= 2/9 and the probability that y recovers but not x is (2/3)(1/3)= 2/9 and so the probability that one but not the other recovers is 2/9+ 2/9= 4/9.

    Thanks for the help, but i still do not understand it.

    So what would be the probabilty
    : that both recover
    : Nither recover
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  4. #4
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    P(A) = 2\left( {\frac{1}{3}} \right)\left( {\frac{2}{3}} \right)\,,\,P(B) = \left( {\frac{1}{3}} \right)^2 \,\& \,P(C) = \left( {\frac{2}{3}} \right)^2
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