An urn contains 5 are red balls and 6 blue balls. Suppose 5 are randomly selected without replacement. What is the probability that exactly 3 are red? Answer correct to four decimal places.

27.94%

Step-by-step explanation:

The statement tells us that we have 5 red balls and 6 blue balls, that is, there are 11 in total (5 + 6)

So the probability of red balls = 5/11 and blue balls probability = 6/11

Let X be the number of red balls of those 5 selected balls.

 Then X follows a binomial distribution with the following parameters:

n = 5

p = 5/11

q = 6/11

P (X) = nCx * p ^ (x) * q ^ (n -x)

required probability is P (X = 3), replacing:

P (X = 3) = 5C3 * (5/11) ^ (3) * (6/11) ^ (5 -3)

P (X = 3) = 5! / (3! (5-3)!) * 0.02794

P (X = 3) = 10 * 0.02794

P (X = 3) = 0.2794

Which means that the probability is 27.94%

3. (5x + 1) - (-10x + 6)

simplify the answer choice. combine like terms. remember to distribute the negative to the terms inside the second parenthesis. also remember that two negatives = one positive, and one of each sign = negative.

- (-10x + 6) = + 10x - 6

simplify. combine like terms:

5x + 1 + 10x - 6

5x + 10x + 1 - 6

15x - 5

∴ it is your answer

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