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Math Help - Probability of getting total score (dice)

  1. #1
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    Probability of getting total score (dice)

    A game is played by players taking turns to toss the dice. If a player gets 2 , 3, 4, or 5 , then it would be his score and the next player takes over. If the score is 1 or 6 , the player is given an extra toss and his score will be the sum of two scores obtained. Event A and B are defined as


    A: The score is 5, 6, 7, 8 or 9
    B: The player tosses twice


    find P (AnB)


    the ans is 1/6 . how to do it?
    can someoene explain it?


    thanks in advance!
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    Re: Probability of getting total score (dice)

    Quote Originally Posted by delso View Post
    A game is played by players taking turns to toss the dice. If a player gets 2 , 3, 4, or 5 , then it would be his score and the next player takes over. If the score is 1 or 6 , the player is given an extra toss and his score will be the sum of two scores obtained. Event A and B are defined as


    A: The score is 5, 6, 7, 8 or 9
    B: The player tosses twice


    find P (AnB)


    the ans is 1/6 . how to do it?
    can someoene explain it?


    thanks in advance!
    there are a few ways to go about this. In this case I think the easiest is to just list the roll sequences that make up $A \cap B$ and find their probabilities.

    $A \cap B$ is pretty clearly $(1,4), (1,5), (1,6), (6,1), (6,2), (6,3)$

    Each event of rolling 2 dice has a probability of $\left(\dfrac 1 6\right)^2=\dfrac 1 {36}$

    There are 6 events in $A \cap B$ so the total probability is $6 \times \dfrac 1 {36} = \dfrac 1 6$
    Thanks from delso
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    Re: Probability of getting total score (dice)

    i am aksed to show that P(A) = 1/3 , but my ans is p(A)= 14/36 , which part is wrong?

    the first toss have 6 options , same as the second toss.

    so my n(total) = 6x6=36Probability of getting total score (dice)-dsc_0045-1-.jpg
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  4. #4
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    Re: Probability of getting total score (dice)

    Quote Originally Posted by delso View Post
    i am aksed to show that P(A) = 1/3 , but my ans is p(A)= 14/36 , which part is wrong?

    the first toss have 6 options , same as the second toss.

    so my n(total) = 6x6=36Click image for larger version. 

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    I showed $A \cap B$ had 6 rolls.

    The only additional roll that is in $A$ but not $B$ is $(5)$

    So I get $Pr[A]=\dfrac 7 {36}$
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