Question

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.)

Answer 1

Answer:

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.