An experiment consisted of 10 draws with replacement from an urn containing four red marbles and six green marbles. The probability that there are at least three and at most six red marbles is:
A)0.563.
B)0.778.
C)0.605.
D)0.82.

Question

Quizplus · Accepted Answer

We can solve this problem using the binomial distribution. Let X be the number of red marbles drawn in ten draws with replacement. Then X ~ B(10, 0.4), since the probability of drawing a red marble is 0.4 (since there are four out of ten marbles that are red).

We want to find P(3 ≤ X ≤ 6), which is the probability that between three and six red marbles are drawn. We can find this probability by adding up the probabilities of drawing exactly three, four, five, or six red marbles:

P(3 ≤ X ≤ 6) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)

Using the binomial probability formula, we can calculate each of these probabilities:

P(X = k) = (10 choose k) * (0.4)^k * (0.6)^(10-k)

where (10 choose k) is the number of ways to choose k red marbles out of ten draws.

Plugging in k = 3, 4, 5, and 6, we get:

P(X = 3) = (10 choose 3) * (0.4)^3 * (0.6)^7 = 0.214
P(X = 4) = (10 choose 4) * (0.4)^4 * (0.6)^6 = 0.326
P(X = 5) = (10 choose 5) * (0.4)^5 * (0.6)^5 = 0.266
P(X = 6) = (10 choose 6) * (0.4)^6 * (0.6)^4 = 0.097

Adding these probabilities together, we get:

P(3 ≤ X ≤ 6) = 0.214 + 0.326 + 0.266 + 0.097 = 0.778

Therefore, the probability that there are at least three and at most six red marbles is 0.778. The best choice is B.

An Experiment Consisted of 10 Draws with Replacement from an Urn