A box contains 5 red and 10 white marbles. If 8 marbles are to be chosen at random (without replacement), determine the probability that (a) 4 will be red, (b) all will be white, (c) at least one will be red.

Using the hypergeometric distribution, it is found that there is a:

a) 0.1632 = 16.32% probability that 4 will be red.

b) 0.0070 = 0.7% probability that all will be white.

c) 0.993 = 99.3% probability that at least one will be red.

The marbles are chosen without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The parameters are:

In this problem:

Item a:

This is P(X = 4), hence:

0.1632 = 16.32% probability that 4 will be red.

Item b:

0 red, hence P(X = 0).

0.0070 = 0.7% probability that all will be white.

Item c:

This is:

From item b:

0.993 = 99.3% probability that at least one will be red.

A similar problem is given at link

a) 0.0023

b) 0.007

c) 0.993

Step-by-step explanation:

The total number of marbles are;

5 + 10 = 15

Probability of red = 5/15

Probability of white = 8/15

Probability of 4 being red means, 4 will be white too;

Thus, the probability without replacement will be;

5/15 * 4/14 * 3/13 * 2/12 * 10/11 * 9/10 * 8/9 * 7/8 =

0.002331002331 which is approximately 0.0023

b. All white

Here, we have all the marbles selected being white

Thus the probability without replacement will be;

10/15 * 9/14 * 8/13 * 7/12 * 6/11 * 5/10 * 4/9 * 3/8

= 0.006993006993 which is approximately 0.007

c. At least one will be red

Mathematically, that will be;

1 - p(no red)

p(no red) = p(all white)

P(all white) = 0.007

hence P(no red) = 0.007

P(at least one red) = 1 - 0.007 = 0.993

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