time for which the product is held in one department has no effect on the time it spends in
any subsequent department. What is the probability that there will be a delay in the manu-
facture of this product?Calculation Procedure:- Compute the probability that the product will be manufactured
without any delay
Since it is certain that the product either will or will not be delayed in a department and
the probability of certainty is 1, probability of no delay = 1 - probability of delay. Thus,
the probability that the product will pass through department B without delay = 1 - 0.15 =
0.85. The probability that the product will pass through every department without delay is
the product of the probabilities of these individual events. Thus, probability of no delay in
manufacture - (0.94)(0.85)(0.97)(0.93)(0.87) = 0.6271. - Compute the probability of a delay in manufacture
Probability of delay in manufacture = 1 - probability of no delay in manufacture = 1 -
0.6271-0.3729.
Related Calculations: This method of calculation can be applied to any situation
where a series of trials occurs, either simultaneously or in sequence, and any trial can
cause the given event. Thus, assume that several projectiles are fired simultaneously and
the probability of landing in a target area is known for each projectile. The above method
can be used to find the probability that at least one projectile will land in the target area.
BINOMIAL PROBABILITYDISTRIBUTION
A case contains 14 units, 9 of which are of type A. Five units will be drawn at random
from the case; and as a unit is drawn, it will be replaced with one of identical type. IfX
denotes the number of type A units drawn, find the probability distribution of X and the
average value of X in the long run.
Calculation Procedure:- Compute the probability corresponding to a particular value
ofX
Consider that n independent trials are performed, and let X denote the number of times an
event E occurs in these n trials. The probability distribution of X is called binomial In this
case, since each unit drawn is replaced with one of identical type, each drawing is inde-
pendent of all preceding drawings; therefore, X has a binomial probability distribution.
The event E consists of drawing a type A unit.
With respect to every drawing, probability of drawing a type A unit = 9/14, and prob-
ability of drawing a unit of some other type = 5/14. Arbitrarily set Jf = 3, and assume that
the units are drawn thus: A-A-A-N-N, where N denotes a type other than A. The proba-
bility of drawing the units in this sequence = (9/14)(9/14)(9/14)(5/14)(5/14) =
(9/14)
3
(5/14)
2
. Clearly this is also the probability of drawing 3 type A units in any other
sequence. Since the type A units can occupy any 3 of the 5 positions in the set of draw-
ings and the exact positions do not matter, the number of sets of drawings that contain 3