type A units = C 53. Summing the probabilities, we find P(3) = C 53 (9/14)^3 (5/14)^2 =
5!/(3!2!)^3 (5/14)^2 = 0.3389.
- Write the equation of binomial probability distribution
Generalize from step 1 to obtain P(X) = Cn^(I - P)n~x where P = probability event E
will occur on a single trial. Here P = 9/14. - Apply the foregoing equation to find the probability
distribution of X
The results are P(O) = 1(9/14)°(5/14)^5 = 0.0058. Similarly, P(Y) = 0.0523; P(2) = 0.1883;
P(3) = 0.3389 from step 1; P(4) = 0.3050; P(S) = 0.109C - Verify the values of probability
Since it is certain that X will assume some value from O to 5, inclusive, the foregoing
probabilities must total 1. There sum is found to be 1.0001, and the results are thus con-
firmed.
5. Compute the average value of X in the long run
Consider that there are an infinite number of cases of the type described and that 5 units
will be drawn from each case in the manner described, thereby generating an infinite set
of values of X. Since the chance of obtaining a type A unit on a single drawing is 9/14, the
average number of type A units that will be obtained in 5 drawings is 5(9/14) = 45/14 =
3.21. Thus, the arithmetic mean of this infinite set of values of X is 3.21.
Alternatively, find the average value of X by multiplying all X values by their respec-
tive probabilities, to get 0.0523 + 2(0.1883) + 3(0.3389) + 4(0.3050) + 5(0.1098) = 3.21.
The arithmetic mean of an infinite set of Jf values is also called the expected value of X.
PASCAL PROBABILITY DISTRIBUTION
Objects are ejected randomly from a rotating mechanism, and the probability that an ob-
ject will enter a stationary receptacle after leaving the mechanism is 0.35. The process of
ejecting objects will continue until four objects have entered the receptacle. Let X denote
the number of objects that must be ejected. Find (a) the probability corresponding to
every X value from 4 to 10, inclusive; (b) the probability that more than 10 objects must
be ejected; (c) the average value of Xin the long run.
Calculation Procedure:- Compute the probability corresponding to a particular
value of X
Consider that a trial is performed repeatedly, each trial being independent of all preceding
trials, until a given event E has occurred for the Mi time. Let X denote the number of tri-
als required. The variable X is said to have a Pascal probability distribution. (In the spe-
cial case where k = 1, the probability distribution is called geometric.) In the present situ-
ation, the given event is entrance of the object into the receptacle, and k = 4.
Use this code: A signifies the object has entered; B signifies it has not. Arbitrarily set
X = 9, and consider this sequence of events: A-B-B-A-B-B-B-A-A, which contains four
A's and five B's. The probability of this sequence, and of every sequence containing four
A's and five B's, is (0.35)
4
(0.65)
5
. Other arrangements corresponding to X= 9 can be ob-
tained by holding the fourth A in the ninth position and rearranging the preceding letters,
which consist of three A's and five B's. Since the A's can be assigned to any 3 of the 8