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mathmath333

  • one year ago

probability question

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{Probability of solving specific problem independently by A and B are}\hspace{.33em}\\~\\ & \dfrac12\ \ \normalsize \text{and}\ \ \dfrac13\ \normalsize \text{respectively.}\hspace{.33em}\\~\\ & \normalsize \text{If both try to solve the problem independently, find the probability}\hspace{.33em}\\~\\ & \normalsize \text{ that exactly one of them solves the problem.}\hspace{.33em}\\~\\ \end{align}}\)

  2. ganeshie8
    • one year ago
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    you want to find below two probabilities and then add them : 1) A solves the problem but B doesn't 2) A doesn't solve the problem but B solves it

  3. ganeshie8
    • one year ago
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    P("A solves" AND "B doesnt solve") = ?

  4. mathmath333
    • one year ago
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    how did u know i need to find this 1) A solves the problem but B doesn't 2) A doesn't solve the problem but B solves it

  5. ganeshie8
    • one year ago
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    good question, thats because of the phrase "exactly one of them"

  6. ganeshie8
    • one year ago
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    "exactly one of them" means only one of them solves the problem

  7. mathmath333
    • one year ago
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    ok

  8. mathmath333
    • one year ago
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    P("A solves" AND "B doesnt solve") = P(A\(\cap\)B')

  9. mathmath333
    • one year ago
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    is this correct P("A solves" AND "B doesnt solve") = P(A∩B')

  10. phi
    • one year ago
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    yes

  11. ganeshie8
    • one year ago
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    Correct, it is easy, go ahead and find it...

  12. mathmath333
    • one year ago
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    P(A∩B')=P(A)-P(A∩B)

  13. ganeshie8
    • one year ago
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    i suggest you not use the formulas to solve these problems, use the problem to make sense of formulas instead

  14. thomas5267
    • one year ago
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    There are only four situations. 1. A solves it and B couldn't. 2. A couldn't solve it and B solves it. 3. A and B both solve it. 4. A and B both couldn't solve it (idiots!) The question is asking for the probability of exactly one of them solves it (or one of them is an idiot!), so it is asking for the probability of 1 or 2 happening.

  15. mathmath333
    • one year ago
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    how can i solve it without formula "P(A∩B')"

  16. ganeshie8
    • one year ago
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    you're given, probability that B solves the problem = 1/3 so can you guess the probability that B couldn't solve the problem ?

  17. mathmath333
    • one year ago
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    the probability that B couldn't solve the problem=2/3

  18. thomas5267
    • one year ago
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    I think it is true for all events X, P(X')=1-P(X).

  19. ganeshie8
    • one year ago
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    Yes, since the events are independent, simply multiply the probabilities : P("A solves" AND "B doesnt solve") = P("A solves")*P("B doesnt solve") = ?

  20. mathmath333
    • one year ago
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    the probability that B couldn't solve the problem =1/2*2/3=1/3

  21. thomas5267
    • one year ago
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    Put it in another way, an event can either happen or not happen. No event can not happen and not not happen at the same time.

  22. ganeshie8
    • one year ago
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    Correct. try finding the probability for other case too : 2) A doesn't solve the problem but B solves it

  23. mathmath333
    • one year ago
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    P("A doesnt solves" AND "B does solve") =1/6

  24. ganeshie8
    • one year ago
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    Add them up and you're done!

  25. mathmath333
    • one year ago
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    =1/2

  26. ganeshie8
    • one year ago
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    1/3 + 1/6 = 3/6 yeah 1/2 looks good to me

  27. mathmath333
    • one year ago
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    thnx

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