An urn contains three white balls and two red balls. The balls are drawn from the urn, oneat a time without replacement, until a white ball is drawn. LetXbe the number of the drawon which a white ball is drawn for the first time. Determine the distribution ofX. Verify it isindeed a pmf. What isE (X)

p (X = 1) = 0.6, p (X = 2) = 0.3, p (X = 3) = 0.1

Verified

E (X) = 1.5

Step-by-step explanation:

Solution:-

- An urn contains the following colored balls:

Color Number of balls

White 3

Red 2

- A ball is drawn from urn without replacement until a white ball is drawn for the first time.

- We will construct cases to determine the distribution of the random-variable X: The number of trials it takes to get the first white ball.

- We have three following case:

1) White ball is drawn on the first attempt (X = 1). The probability of drawing a white ball in the first trial would be:

p (X = 1) = (Number of white balls) / (Total number of ball)

p (X = 1) = (3) / (5)

2) A red ball is drawn on the first draw and a white ball is drawn on the second trial (X = 2). The probability of drawing a red ball first would be:

p (Red on first trial) = (Number of red balls) / (Total number of balls)

p (Red on first trial) = (2) / (5)

- Then draw a white ball from a total of 4 balls left in the urn (remember without replacement).

p (White on second trial) = (Number of white balls) / (number of balls left)

p (White on second trial) = (3) / (4)

- Then to draw red on first trial and white ball on second trial we can express:

p (X = 2) = p (Red on first trial) * p (White on second trial)

p (X = 2) = (2 / 5) * (3 / 4)

p (X = 2) = (3 / 10)

3) A red ball is drawn on the first draw and second draw and then a white ball is drawn on the third trial (X = 3). The probability of drawing a red ball first would be (2 / 5). Then we are left with 4 balls in the urn, we again draw a red ball:

p (Red on second trial) = (Number of red balls) / (number of balls left)

p (Red on second trial) = (1) / (4)

- Then draw a white ball from a total of 3 balls left in the urn (remember without replacement).

p (White on 3rd trial) = (Number of white balls) / (number of balls left)

p (White on 3rd trial) = (3) / (3) = 1

- Then to draw red on first two trials and white ball on third trial we can express:

p (X = 3) = p (Red on 1st trial) * p (Red on 2nd trial) * p (White on 3rd trial)

p (X = 3) = (2 / 5) * (1 / 4) * 1

p (X = 3) = (1 / 10)

- The probability distribution of X is as follows:

X 1 2 3

p (X) 0.6 0.3 0.1

- To verify the above the distribution. We will sum all the probabilities for all outcomes (X = 1, 2, 3) must be equal to 1.

∑ p (Xi) = 0.6 + 0.3 + 0.1

= 1 (proven it is indeed a pmf)

- The expected value E (X) of the distribution i. e the expected number of trials until we draw a white ball for the first time:

E (X) = ∑ [ p (Xi) * Xi ]

E (X) = (1) * (0.6) + (2) * (0.3) + (3) * (0.1)

E (X) = 0.6 + 0.6 + 0.3

E (X) = 1.5 trials until first white ball is drawn.