Application of Probability to Mechanical Design
43
and
P(B) = P(A)P(B/A) + P(f4)P(B/)4)
(2.8)
placing Eq. (2.8) into Eq. (2.7)
P(A)P(B/A)
P(A/B) = P(A)P(B/A) + P(f4)P(B/)4)
(2.9)
for "A" with more than two alternates
P(A)P(B/A)
P(A/B) =
(2.10)
}-]4 P(Ai)P(B/
EXAMPLE 2.4. [2.17]. Using the Example 2.3 table for E1 and E2
find P(Ez/EI) using Eq. (2.5). The probability of "El" has happened, that
"E2" will follow, note "E2" happens in nl and n3 but "El" only occurs
in nl so
nl
_ nl
n
P(E1E2)
P(Ez/E1) nl + n2 nl + n2 - P(EI)
should El and E2 be independent which means El and E2 can happen separately or it means when El occurs E~ does not follow.
now
P(E~ E2) = P(E~)P(E~)
nl + n2
P(El) - -- -- 2/4
nl ÷ n3
P(E2) - -- -- 2/4
4 1
nl
P(E1E2) = (2/4)(2/4) = 1V = ~ or
if the events are not independent
P(E1 E2) = P(E2)P(E2/El)
EXAMPLE 2.5. A sorting example is solved using Eq. (2.8).
Given are two urns in a box, urn 1 with 3 white balls and 5 red balls;
urn 2 with 5 white balls and 7 red balls.