Question

There are 6 yellow, 9 blue, and 4 orange balls inside an urn. 8 balls are chosen from the urn without replacement. Let X, Y , and W denote the number of yellow, blue, and orange balls selected, respectively.

(a) Find P{X + W = 1}.
(b) Find P{X = 2, Y = W}

Answer 1

Using the probability concept, it is found that:

a) P{X + W = 1} = 0

b) P{X = 2, Y = W} = 0.0667

A probability is the number of desired outcomes divided by the number of total outcomes.

In this problem, the order in which the balls are chosen is not important, hence, the combination formula is used to find the number of outcomes.

Combination formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by:

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In total, 8 balls are chosen from a set of 19, hence:

$T = C_{19,8} = \frac{19!}{8!11!} = 75582$

Item a:

• 1 from a set of 15.
• 7 from a set of 4.

Not possible, hence:

P{X + W = 1} = 0

Item b:

• 2 from a set of 6.
• 3 from a set of 9.
• 3 from a set of 4.

Hence:

$D = C_{6,2}C_{9,3}C_{4,3} = \frac{6!}{2!4!} \times \frac{9!}{3!6!} \times \frac{4!}{3!1!} = 5040$

Then, the probability is:

$p = \frac{D}{T} = \frac{5040}{75582} = 0.0667$

P{X = 2, Y = W} = 0.0667

A similar problem is given at https://brainly.com/question/15536019