Question

) Suppose that an urn contains 7 green, 8 red, and 9 yellow balls. You pick a ball at random and replace it. You then do this 9 more times (making 10 times altogether). What is the probability that you selected green 2 times, red 3 times, and yellow 5 times?

Answer 1

Answer:

Probability  = 0.0589

Step-by-step explanation:

The total number of balls in an urn is: 7 + 8 + 9 = 24 balls

The possible arrangements are: $\frac{10!}{2!3!5!} = 2520$

Probability of getting a green ball = $\frac{7}{24}$.

Probability of getting a red ball = $\frac{8}{24}$.

Probability of getting a yellow ball = .

Therefore, the required probability is:

$2520 \times (\frac{7}{24})^2 \times (\frac{8}{24})^3 \times (\frac{9}{24})^5\\ = 2520 \times 0.08507 \times 0.03704 \times 0.00742 \\= 0.0589$

Answer 2

Title:

The required probability is $2.336\times 10^{-5} = 0.00002$.

Step-by-step explanation:

In the urn, there are total (7 + 8 + 9) = 24 balls.

We are taking 10 balls out of the urn with replacement.

In each tern, we have overall 10 choices to chose 1 ball.

If we want to get a green, there are 7 choices each time.

Similarly, 8 choices each time for getting a red and 9 choices each time for getting a yellow ball.

The probability of getting green 2 times, red 3 times, and yellow 5 times is $\frac{7^2 \times8^3\times9^5}{(24)^{10}} = 2.336 \times 10^{-5}$.