Two fair dice are rolled 3 times and the sums of the numbers that come up is recorded. Find the probability of these events. A) the sum 10 on each of the three rolls? B) the sum is 10 exactly twice in three rolls?

A) P(10 on three rolls) = 1/1728 = 0.0005787

B) P(10 in two of three rolls) = 11/1728 = 0.0063657

Step-by-step explanation:

Each die has 6 possible values, so a pair of dice have a total of 6*6=36 outcomes.

To find a sum of 10, the cases are:

(4,6), (5,5) and (6,4).

So to have a sum of 10 we have 3 cases among the 36 possible, therefore the probability is:

P(10) = 3/36 = 1/12

A)

If we want the sum of 10 in each of the three rolls, the probability is:

P(10 on three rolls) = P(10)^3 = (1/12)^3 = 1/1728 = 0.0005787

B)

If the sum is 10 in two of three rolls, in one roll we need the probability of the sum not being 10:

P(not 10) = 1 - P(10) = 1 - (1/12) = 11/12

So we have:

P(10 in two of three rolls) = P(10)^2 * P(not 10) = (1/12)^2 * (11/12)

P(10 in two of three rolls) = 11/1728 = 0.0063657

4 + 5 7/9= 9 7/9

step-by-step explanation:

answer: the price of one pound cabbage = $ 0.8

the point (10,8) represents that the cost of 10 pounds cabbage is $ 8

step-by-step explanation:

since, here is a proportional relation between the number of pounds of cabbage (x) and the price in dollars (y).

therefore, we can write,

⇒ y = k x

where k is the constant of proportionality.

now, point (5,4) is in the line.

thus, it will satisfies the above equation of line.

⇒ 4 = 5 k

⇒ k = 0.8

thus, the equation of the line is,

y = 0.8 x

a) for x = 1,

y = 0.8

therefore, the cost of one pound of cabbage is 0.8 dollars.

b) in the below graph (10,8) represents that the cost of 10 pounds of cabbage is 8 dollars.