Question

Can somebody help me with these. with solution

1 Two balls are drawn in succession without replacement from a box containing 4 red balls and 3 black balls. Let Y be the random variable, where Y is the number of the red balls. What is the expected value of this random event?

2. Two balls are drawn in succession from a box without repetition containing 3 red 3 blue and four yellow marbles. What is the probability that they are of the same color?

Answer 1

See below, please.

Step-by-step explanation:

Part 1.

Let Y be the random variable representing the number of red balls drawn in succession. The possible outcomes are (R, R), (R, B), (B, R), and (B, B). We can calculate the probability of each outcome as follows

P(Y=0) = P(B, B) = (3/7) * (2/6) = 1/7

P(Y=1) = P(R, B) + P(B, R) = (4/7) * (3/6) + (3/7) * (4/6) = 12/42 + 12/42 = 24/42

P(Y=2) = P(R, R) = (4/7) * (3/6) = 2/7

The expected value of Y is given by:

E(Y) = Σ yi * P(Y=y)

= 0 * (1/7) + 1 * (24/42) + 2 * (2/7)

= 0 + 8/14 + 4/7

= 16/14

= 1.14

Therefore, the expected value of Y is 1.14.

Part 2.

The total number of ways to draw two marbles from the box is 10C2 = 45. The number of ways to draw two marbles of the same color is 3C2 + 4C2 = 3 + 6 = 9. Therefore, the probability of drawing two marbles of the same color is

P(same color) = 9/45 = 1/5

Therefore, the probability that the two balls are of the same color is 1/5.