. An urn contains six colored balls: four orange and two blue. Two balls are selected at random without replacement, and you are told that at least one of them is orange. What is the probability that the other ball is also orange?

We draw two balls. We know that one of them is orange.

Let's suppose that the first one is the orange.

Then after drawing that, we know that there are 5 balls left in the urn, and only one will be orange.

Then the probability of drawing the other orange ball is equal to the quotient between the number of orange balls and the total number of balls.

P1 = 1/5.

And now let's consider the other case, where we know that in the second draw we will draw an orange ball.

Then the probability of drawing an orange ball in the first draw is equal to the quotient between the number of orange balls (at the beginning we have 2) and the total number of balls (6)

P2 = 2/6

As those represent different cases (we assume different conditions for each one), the probability that we draw two orange balls, knowing for sure that we will draw one, is equal to the sum of these probabilities.

P = P1 + P2 = 1/5 + 2/6 = 6/30 + 10/30 = 16/30 = 8/15

2

step-by-2step 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.