# Thread: Need help with the second part of this question.

1. ## Need help with the second part of this question.

A six-sided die is weighted so that the probability of rolling a 1 is twice the probability of rolling any other number. what is the probability that the sum of two numbers thrown in two tries is equal to 4?

here is how far i got:

we are looking for the probability of rolling (3,1) (2,2) or (1,3) with this die.

Probability of rolling a 1 is 2/7 and probability of rolling any other number is 1/7

I do not know how to finish the problem from here.

2. Originally Posted by bm203
A six-sided die is weighted so that the probability of rolling a 1 is twice the probability of rolling any other number. what is the probability that the sum of two numbers thrown in two tries is equal to 4?

here is how far i got:

we are looking for the probability of rolling (3,1) (2,2) or (1,3) with this die.

Probability of rolling a 1 is 2/7 and probability of rolling any other number is 1/7

I do not know how to finish the problem from here.
With these probabilities

$p[1,3]=\frac{2}{7}\frac{1}{7}$

$p[3,1]=\frac{1}{7}\frac{2}{7}$

$p[2,2]=\frac{1}{7}\frac{1}{7}$

sum the probabilities as these are the only ways to score a combined 4.