Consider two events A and E. We can write A as the union of two mutually exclusive events, (E ∩ A) and (E' ∩ A). Hence A = (E ∩ A) U (E' ∩ A).
By the property of mutually exclusive events,
P(A) = P[(E ∩ A) U (E' ∩ A)]
= P(E ∩ A) + P(E' ∩ A)
= P(E).P(A/E) + P(E').P(A/E') ..... by definition of conditional probability
Now the above can be extended if the events B1, B2,......,Bk constitute a partition of the sample space, S such that P(Bi) ≠ 0, then for any event A (shown in yellow) of S, we have
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P(A) = åi=1k [P(Bi ∩ A)] = åi=1k [P(Bi) . P(A/Bi)] |
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Example: In a certain assembly plant, three machines B1, B2 and B3 make 30%, 45% and 25% of the products, respectively. About 2%, 3% and 2% of the products made by each machine, respectively, are defective. Now if a finished product is selected randomly, what is the probability that it is defective?
Answer: Let event A = the product selected is defective.
Let B1= the product was made by machine B1.
Let B2= the product was made by machine B2.
Let B3= the product was made by machine B3.
So we have P(A) = P(B1).P(A/B1) + P(B2).P(A/B2) + P(B3).P(A/B3)
= (0.3)(0.02) + (0.45)(0.03) + (0.25)(0.02)
= 0.006 + 0.0135 + 0.005
= 0.0245
Now if we wish to find the conditional P(B1/A) in the above example, i.e. given that the product selected is defective, which one of the machines made it. This can be found by using the Bayes' rule.
Bayes' Rule: If the events B1, B2,......,Bk constitute a partition of the sample space, S such that P(Bi) ≠ 0, then for any event A of S, such that P(A) ≠ 0,
P(Br/A) = P(Br ∩ A) = P(Br).P(A/Br) ... for r = 1 to k
åi=1k [P(Bi ∩ A)] åi=1k [P(Br).P(A/Br)]
Example: If for the above problem, the product chosen randomly was found to be defective, what is the probability that it was made by machine B3?
Answer: P(B3/A) = [P(B3).P(A/B3)] / [ P(B1).P(A/B1) + P(B2).P(A/B2) + P(B3).P(A/B3) ]
= [(0.25)(0.02)] / [ (0.3)(0.02) + (0.45)(0.03) + (0.25)(0.02) ]
= 0.005 / 0.0245
= 0.2041
A regional telephone company operates three identical relay stations, A B C at different locations. During one year the problems reported by each station along with their causes is shown below.
| A | B | C | |
| Problems with electricity supply | 2 | 4 | 1 |
| Computer malfunction | 7 | 3 | 2 |
| Malfunctioning electrical equipment | 1 | 4 | 4 |
| Caused by human errors | 7 | 5 | 6 |
If a malfunction caused by human error is reported, find the probability that it came from station C.
The city police enforce speed limits by using radar traps at 4 different locations, L1 L2 L3 and L4. They are operated 40%, 30% , 20% and 10% of the time, respectively. If a speeding person, who has a probability of 0.2, 0.3, 0.1, and 0.4 of passing through these locations, what is the probability that he will receive a ticket? Given that he has got a ticket, what is the most probable location where he was caught speeding?
