Application of Probability to Mechanical Design
41
For a case of three events A + B + C let X = A’ + B’ for A in Eq. (2.2) then
P(A + B + C) = P(X + C) = P(X) + P(C)
= P(X) + P(C) - P(X)P(C)
now substitute
X = A I + B’
P(x) = P(A’ + B’) + P(C) - P(C)[P(A’ B’)]
now use Eq. (2.2) for P(A’+
P(A + B + C) = P(A’) + P(B’) + P(C) - P(A’B’) -
+ P(B’) - P(A’B’)I
Also it is found
P(AIB) = P(AI)P(B ~) for independent events which is substituted into
the equation
P(A + B + C) = [P(A’) P(B’) - P(A’)P(B’)] +
- P(C)] [P(A’) + P(B’) P(A’)P(B’)]
= P(A’) + P(B’) + P(C) - P(A’)P(B’) -
- e(c)e(B’) ~’ (A’)~’(B’)t’(C)
dropping the primes yields
P(A + B ÷ C) = P(A) + P(B) + P(C) - P(A)P(B)
(2.3)
- P(C)P(B) + P(A)P(B)P(C)
EXAMPLE 2.3 [2.17]. Discrete events E1 or E2 may be approached
using Example 2.2, (/~1 or/~2 means it doesn’t occur).
nl
n2
n3
n4
n
E~ E~
E~ E~
E~ E~
E~ E2
Total
P(E1) nl + n2 P(E 2) - nl ÷ n~3
n
n
Now consider P(E~ + E2) probability of E1 or E2 or both
n~ + n2 -b n3
P(E~ + E2)
n
Now to substitute for P(EI) and P(E2) which are sums yielding
2n~+n2+n3
n~
P(E~ + E2) = P(E1) + P(E2) -- P(EIE2)
n
~/