Apair of dice is loaded. the probability that a 4 appears on the first die is 2/7, and the probability that a 3 appears on the second die is 2/7. other outcomes for each die appear with probability 1/7. what is the probability of 7 appearing as the sum of the numbers when the two dice are rolled?

9/49

Step-by-step explanation:

For the first die, probability of obtaining 4= 2/7

For the second die, the probability of obtaining 3= 2/7

Other outcomes for each die appear with a probability of 1/7

There are six cases of obtaining the sum of the two die as 7.

E1 = 1 and 6

E2 = 2 and 5

E3 = 3 and 4

E4 = 4 and 3

E5 = 5 and 2

E6 = 6 and 1

Pr(E1) = 1/7*1/7

= 1/49

Pr(E2) = 1/7*1/7

= 1/49

Pr(E3) = 1/7*1/7

= 1/49

Pr(E4) = 2/7*2/7

= 4/49

Pr(E5) = 1/7*1/7

= 1/49

Pr(E6)= 1/7*1/7

= 1/49

P(E) = Pr(E1) + Pr(E2) + Pr(E3) + Pr(E4) + Pr(E5) + Pr(E6)

= 1/49 + 1/49 + 1/49 + 4/49 + 1/49 + 1/49

= 9/49

a

step-by-step explanation: