Question

There are six red balls and three green balls in a box. If we randomly select 3 balls from the box with replacement, how likely do we observe exactly one green ball?

Select one:
a.The probability is between 0.25 and 0.35.
b.The probability is between 0.15 and 0.25.
c.The probability is between 0.00 and 0.15.
d.The probability is between 0.45 and 0.60.
e. The probability is between 0.35 and 0.45.

Answer 1

The probability of selecting exactly one green ball from a box with replacement, after three picks, is approximately 0.44. Using the binomial probability formula, this falls within the range of 0.35 to 0.45, and therefore, option e is correct.

Step-by-step explanation:

Calculating the Probability of Selecting Exactly One Green Ball

To find the probability of observing exactly one green ball when selecting 3 balls from the box with replacement, we can use the binomial probability formula. Since the balls are being replaced, the probability of picking a green ball (success) remains constant at 3/9 or 1/3 for every pick, and the probability of picking a red ball (failure) remains constant at 6/9 or 2/3.

The binomial probability formula is:

P(X = k) = (n choose k) *
`(p^k) * (q^{(n-k))`

Where:

n is the number of trials (3 balls are picked)

k is the number of successes (1 green ball)

p is the probability of success on a single trial (1/3)

q is the probability of failure on a single trial (2/3)

Substituting the values we get:

P(X = 1) = (3 choose 1) *
`(1/3)^1 * (2/3)^2`

This simplifies to:

P(X = 1) = 3 * (1/3) * (4/9)

P(X = 1) = 4/9 or approximately 0.44

Therefore, the probability of selecting exactly one green ball is between 0.35 and 0.45, making option e the correct answer.