# Math Help - Probability of getting total score (dice)

1. ## 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!

2. ## Re: Probability of getting total score (dice)

Originally Posted by delso
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 $6 \times \dfrac 1 {36} = \dfrac 1 6$

3. ## 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=36

4. ## Re: Probability of getting total score (dice)

Originally Posted by delso
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=36
I showed $A \cap B$ had 6 rolls.

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

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