Bayes’ Theorem
Let B1, B2, B3, …, Bn be n pairwise mutually exclusive and exhaustive set of events connected to a random experiment E where at least one of B1, B2, B3, …, Bn is sure to happen. Let A be an arbitrary event connected to E, where P(A) is not zero. Then,
\[P({{B}_{i}}|A)=\frac{P({{B}_{i}}).P(A|{{B}_{i}})}{\sum\limits_{i=1}^{n}{[P({{B}_{i}}).P(A|{{B}_{i}})]}}\]
Proof:
Here, \[P(A),P({{B}_{i}}),P(A|{{B}_{i}})>0\]
Then we can represent P(A|Bi) in two ways,
\[P(A\cap {{B}_{i}})=P({{B}_{i}}).P(A|{{B}_{i}})…………….(i)\]
\[P(A\cap {{B}_{i}})=P(A).P({{B}_{i}}|A)…………….(ii)\]
Now from (i) and (ii) we get,
\[P({{B}_{i}}).P(A|{{B}_{i}})=P(A).P({{B}_{i}}|A)\]
\[\therefore P({{B}_{i}}|A)=\frac{P({{B}_{i}}).P(A|{{B}_{i}})}{P(A)}…………..(iii)\]
Let, S be the sample space of E. Now,
\[A=SA=({{B}_{1}}+{{B}_{2}}+{{B}_{3}}+…+{{B}_{n}})A\]
\[=({{B}_{1}}A+{{B}_{2}}A+{{B}_{3}}A+…+{{B}_{n}}A)\]
Again,
\[({{B}_{i}}A)({{B}_{j}}A)={{B}_{i}}{{B}_{j}}A=\phi A=\phi (i\ne j)\]
So, B1A, B2A, B3A, …, BnA are mutually exclusive
\[\therefore P(A)=\sum\limits_{i=1}^{n}{P(A\cap {{B}_{i}})}\]
\[=\sum\limits_{i=1}^{n}{P({{B}_{i}})}.P(A|{{B}_{i}})\]
[Using law of Total Probability]
Now from (iii) we get,
\[P({{B}_{i}}|A)=\frac{P({{B}_{i}}).P(A|{{B}_{i}})}{\sum\limits_{i=1}^{n}{[P({{B}_{i}}).P(A|{{B}_{i}})]}}\]
The bag A has 3 white and 2 red balls and another bag B has 4 white and 5 red balls. A bag is chosen randomly and a ball also picked from that bag and seen that the ball picked is red. Find the probability that the bag chosen was B.
Solution:
Let, the events E1 = ‘Chosen bag is A’, E2 = ‘Chosen bag is B’ and E = ‘red ball picked’.
Since, Picked ball is red so we have to find P(E2|E).
\[P({{E}_{1}})=\frac{2}{2}=P({{E}_{2}})\]
Again, probability of ‘bag A is chosen and red ball picked’ is
\[P(E|{{E}_{1}})=\frac{2}{5}\]
Similarly,
\[P(E|{{E}_{2}})=\frac{5}{9}\]
Now, from Bayes’ Theorem we have,
\[P({{E}_{2}}|E)=\frac{P({{E}_{2}}).P(E|{{E}_{2}})}{P({{E}_{1}}).P(E|{{E}_{1}})+P({{E}_{2}}).P(E|{{E}_{2}})}\]
\[=\frac{\frac{1}{2}.\frac{5}{9}}{\frac{1}{2}.\frac{2}{5}+\frac{1}{2}.\frac{5}{9}}\]
\[=\frac{\frac{5}{18}}{\frac{1}{5}+\frac{5}{18}}\]
\[=\frac{\frac{5}{18}}{\frac{18+25}{90}}\]
\[=\frac{5}{18}.\frac{90}{43}\]
\[=\frac{25}{43}\]
A company has two plants to manufacture scooters. Plant X manufactures 70% of scooters and plant Y manufactures 30%. At plant X, 80% of scooters are rated standard quality and at plant Y, 90% of scooters are rated standard quality. A scooter is picked up at random and is found to be of standard quality. What is the chance that it has come from plant X, plant Y?
Let us define the following events:
H1: scooter is manufactured by plant X
H2: scooter is manufactured by plant Y
A: scooter is rated as standard quality
Then we are given,
\[P({{H}_{1}})=\frac{70}{100}=0.70\]
\[P({{H}_{2}})=\frac{30}{100}=0.30\]
\[P(A/{{H}_{1}})=\frac{80}{100}=0.80\]
\[P(A/{{H}_{2}})=\frac{90}{100}=0.90\]
The probability that scooter comes from plant X is of standard quality is,
\[P({{H}_{1}}/A)=\frac{P({{H}_{1}}).P(A/{{H}_{1}})}{P({{H}_{1}}).P(A/{{H}_{1}})+P({{H}_{2}}).P(A/{{H}_{2}})}\]
\[=\frac{(0.70).(0.80)}{(0.70).(0.80)+(0.30).(0.90)}\]
\[=\frac{56}{83}\]
Similarly,
\[P({{H}_{2}}/A)=\frac{P({{H}_{2}}).P(A/{{H}_{2}})}{P({{H}_{1}}).P(A/{{H}_{1}})+P({{H}_{2}}).P(A/{{H}_{2}})}\]
\[=\frac{(0.30).(0.90)}{(0.70).(0.80)+(0.30).(0.90)}\]
\[=\frac{27}{83}\]
Consider the clinical test described at the start of this section. Suppose that 1 in 1000 of the population is a carrier of the disease. Suppose also that the probability that a carrier tests negative is 1%, while the probability that a no carrier tests positive is 5%. (A test achieving these values would be regarded as very successful). i)A patient has just had a positive test result. What is the probability that the patient is a carrier? ii)A patient has just had a negative test result. What is the probability that the patient is a carrier?
Let A be the event ‘the patient is a carrier’, and B the event ‘the test result is positive’. We are given that P(A) = 0.001 (so that P(Ac) = 0.999), and that
\[P(B|A)=0.99\]
\[P(B|{{A}^{c}})=0.05\]
i)
\[P(A/B)=\frac{P(A).P(B/A)}{P(A).P(B/A)+P({{A}^{c}}).P(B/{{A}^{c}})}\]
\[=\frac{\text{(0}\text{.001 }\!\!\times\!\!\text{ }0.99\text{)}}{\text{(0}\text{.001 }\!\!\times\!\!\text{ 0}\text{.99)}+\text{(0}\text{.999 }\!\!\times\!\!\text{ 0}\text{.05)}}\]
\[\text{=0}\text{.0194}\]
ii)
\[P(A/{{B}^{c}})=\frac{P(A).P({{B}^{c}}/A)}{P(A).P({{B}^{c}}/A)+P({{A}^{c}}).P({{B}^{c}}/{{A}^{c}})}\]
\[=\frac{\text{(0}\text{.001 }\!\!\times\!\!\text{ }0.001\text{)}}{\text{(0}\text{.001 }\!\!\times\!\!\text{ }0.001\text{)}+\text{(0}\text{.999 }\!\!\times\!\!\text{ 0}\text{.95)}}\]
\[\text{=0}\text{.00001}\]
2% of the population has a certain blood disease in a serious form; 10% have it in a mild form; and 88% don’t have it at all. A new blood test is developed; the probability of testing positive is 9/10 if the subject has the serious form, 6/10 if the subject has the mild form, and 1/10 if the subject doesn’t have the disease.
I have just tested positive. What is the probability that I have the serious form of the disease?
Let X be ‘has disease in serious form’, Y be ‘has disease in mild form’, and Z be ‘doesn’t have disease’. Let B be ‘test positive’. Then we are given that X, Y, Z form a partition.
P(X) = 0.02, P(Y) = 0.1, P(Z) = 0.88
P(B | X)= 0.9, P(B | Y) = 0.6, P(B | Z) = 0.1
Thus by the theorem of Total Probability,
\[\text{P(B) = 0}\text{.9 }\!\!\times\!\!\text{ 0}\text{.02 +0}\text{.6 }\!\!\times\!\!\text{ 0}\text{.1 +0}\text{.1 }\!\!\times\!\!\text{ 0}\text{.88 = 0}\text{.166,}\]
and then by Bayes’ Theorem,
\[\text{P(X }\!\!|\!\!\text{ B)=}\frac{P(X).P(B|X)}{P(B)}\]
\[=\frac{0.02\text{ }\!\!\times\!\!\text{ }0.9}{0.166}=0.108\]
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