Random Signals for Engineers using MATLAB and Mathcad
Copyright  1999 Springer Verlag NY
Example 1.9
In this example we will calculate the probability of drawing 4 red balls from a bin that contains 4 red balls
and 46 white balls. We will draw the balls assuming that the individual draws are independent and then
recalculate the probabilities when the draws are not. Replacing the ball in the bin every time a draw is made
can simulate the independent case.
SOLUTION: Independent Case -The probability of drawing 4 red balls is just the product of 4 draws where
the P[a red ball is drawn] = Pr .
Pr=4/50;
Since there are 4 red balls available for each draw from the 50 total in the bin, the probability of drawing a
red ball is P[A] and the probability of drawing 4 red balls on 4 tries is
PA1  A2  A3  A4   PA1   PA2   PA3   PA4 
where Ai is the event that represents drawing a red ball. If the ball is replaced for the next draw, when a
new ball is drawn the conditions are the same as the previous draw. We have an independent draw since
the present draw does not depend on the previous one. The P[four red balls in 4 draws] is
Pr^4
ans =
4.0960e-005
Dependent Case - If we do not replace the balls we will use the laws of conditioned probabilities to
compute the P[four red balls in 4 draws].

P A
 
 A   P A
   
 
 A   P A A  A   P A A   P A 
  
P A1  A2  A3  A4  P A2  A3  A4 A1  P A1  P A3  A4 A1  A2  P A2 A1  P A1
1
 A2  A3
4
4
A1  A2
3
3
1
2
2
1
1
This equation states that the second draw is conditioned on the first and the third is conditioned on the first
two and the forth is conditioned on the preceeding three. P[A1] is Pr ; P[A2|A1] is Pr1;
P[A3|A1 A2] = Pr2 and P[A4|A1 A2  A3] = Pr3. The probabilities are computed by ratio of ways we
can select a red ball over the number of ball in the bin when each red ball is drawn.
Pr1=3/49;
Pr2=2/48; Pr3=1/47;
In all the cases we reduce the red ball count by 1 and the number in the bin by 1 as we draw red balls out of
the bin. The probability of 4 draws without replacement is
Pr*Pr1*Pr2*Pr3
ans =
4.3422e-006
The values are not the same. For the conditioned case, the probability of drawing a red ball decreases as
we draw red balls and thus the overall probability of drawing 4 red balls conditioned draw is less than the
independent-draw case.