Two balls are selected at random from an urn that contains five white balls and eight red balls. Let the random variable X denote the number of white balls drawn times the number of red balls drawn. Find the probability distribution. (Order your answers from smallest to largest x-value. Round your answers to four decimal places.)

P (X=0) = 19/39 = 0.4872

P (X=1) = 20/39 = 0.5128

Step-by-step explanation:

The goal is find the distribution for the random variable X = "number of white balls drawn times the number of red balls drawn".

Notation.

Let W = White balls and R = red balls

Total number of balls = 5 W + 8R = 13 balls

They selected 2 balls from 13 in total.

Total outcomes

The total number of outcomes to select the 2 balls from a total of 13 are n (sample space) = 13C2 = 13! / (11! * 2!) = 78.

Definition of the random variable X

Let a = number of W balls and b = number of R balls selected on the extraction of the two balls

They selected two balls, so the random variable X would be given by this expression, X = ab.

We identify the possible cases for the pair (a, b), given by:

(0,2), (1,1), (2,0)

The possible values for X are then:

0*2 = 0, 1*1=1, 2*0=0

As we can see X = 0,1.

Calculation of probabilities

The probability for the two possible values for X are:

For the calculations we use the definition of combination, given by:

nCx = (n!) / /[ (n-x) ! * x!]

Calculations

P (X=0) = P[ (0,2) or (2,0) ] = Possible outcomes / total outcomes

= (5C2 + 8C2) / (13C2) = [5! / (3!*2!) + 8! / (6!*2!) ] / [13! / (11!*2!) ]

= (10+28) / 78 = 38/78 = (38/2) / (78/2) = 19/39 = 0.4872 (rounded)

P (x=1) = P[ (1,1) ] = [5C1 * 8C1] / (13C2) = [5! / (4!*1!) + 8! / (7!*1!) ]/[13! / (11!*2!) ]

= 40/78 = (40/2) / (78/2) = 20/39 = 0.5128 (rounded)

And since the sum for the two possible probabilities on the sample space is 1, because 19/39 + 20/39 = 1, we proof that we have a probability distribution.