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%

add all the numbers and check the ones place.

the answer is 486. as the ones place is more than 4, the estimation will be the next 10 ie, 490

the answer is 489

step-by-step explanation: