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 <em>random variable</em> X= "number of white balls drawn times the number of red balls drawn".
<em>Notation.</em>
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.
<em>Total outcomes</em>
The total number of outcomes to select the 2 balls from a total of 13 are n(sample space)= 13C2 = 13!/(11! *2!) = 78.
<em>Definition of the random variable X</em>
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.
<em>Calculation of probabilities</em>
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!]
<em>Calculations</em>
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.
