Page 25 - 4660
P. 25
The Total Probability Formula
Example 3.3. A lot of 100 items undergoes a selective inspection. The entire
lot us rejected if there is at least one defective item if five items checked.
What is the probability that the given lot will be rejected if it contains 5%
defective items? ,
Solution. Find the probability q of the complementary event A consisting of the situation in
which the lot will be accepted. The given event is an intersection of five events
A = A 1 A 2 A 3 A 4 A 5 , where A k (k = 1, 2, 3, 4, 5) means that the k-th item checked is good.
The probability of the event A 1 is P(A 1 ) = 95 since there are only 100 items, of which 95 are
100
good. After the occurrence of the event A 1 , there remain 99 items, of which 94 are good. Thus,
P(A 2 /A 1 ) = 94 . Analogously, P(A 2 /A 1 A 2 ) = 93 , P(A 4 /A 1 A 2 A 3 ) = 92 , P(A 5 /A 1 A 2 A 3 A 4 ) = 91 .
99 98 97 96
According to the formula (3.5), we find that q = 95 · 94 93 92 91 = 0.77 The required probability
·
·
·
100 99 98 97 96
p = 1 − q = 0.23.
The Total Probability Formula
Suppose that an event A can occurs only together with one of events H 1 , H 2 , . . . , H n , which form
a total group.
Events H 1 , H 2 , . . . , H n are called hypotheses.
(A),
Let probabilities P(H 1 ), P(H 2 ), . . . , P(H n ) are known and conditional probabilities P H i
(i = 1, 2, . . . , n) also are known. Find the probability P(A).
Since the events H 1 , H 2 , . . . , H n form a total group their union (sum) is a certain event.
The event A may appear only together with some event H i .
Thus the event A is the union of the events AH 1 , AH 2 , . . . , AH n . As by condition the events H 1 ,
H 2 , . . . , H n are exclusive (incompatible) the events are also exclusive and we can use the addition
theorem
n
∑
P(A) = P(AH 1 ) + P(AH 2 ) + . . . + P(AH n ) = P(AH k ).
k=1
On using the product theorem we obtain
(A)
P(AH k ) = P(H k )P H k
and therefore
n
∑
P(A) = P(H k )P H k (A). (3.7)
k=1
The equation (3.7) is called a Total Probability Formula.
Example 3.4. Electric bulbs are produced at two plants. Products of the first
plant contains 70% standard bulbs, the second – 80%. The bulbs were sent for sale
to shops. To the first shop were sent 60% of the general amount of bulbs, to the
second – 40%.
What is the probability, that at random purchased bulb at the shop is a standard
bulb? ,
Solution. Denote by A the event meaning that a standard bulb was bought, by H 1 — a bulb
was produced at the first plant, by H 2 — a bulb was produced at the second plant.
(A) = 0.8.
We have P(H 1 ) = 0.6, P(H 2 ) = 0.4, P H 1 (A) = 0.7, P H 2
By the total probability formula we have
(A) = 0.6 · 0.7 + 0.4 · 0.8 = 0.74.
P(A) = P(H 1 ) · P H 1 (A) + P(H 2 ) · P H 2
25