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Bayes' Rule
Submitted By
Md. Isarot Hossan
ID: 142014001
Department of Textile Engineering
Green University of Bangladesh
Bayes' rule
 True Bayesians actually consider conditional probabilities as
more basic than joint probabilities. It is easy to define
P(A|B) without reference to the joint probability P(A,B). To
see this note that we can rearrange the conditional
probability formula to get:
P(A|B) P(B) = P(A,B)
Bayes' Rule
But by symmetry we can also get:
P(B|A) P(A) = P(A,B)
It follows that:
Or,
Which is the so-called Bayes' Rule.
Bayes Rule Example
 Suppose that we have two bags each containing black and white balls.
One bag contains three times as many white balls as blacks. The other
bag contains three times as many black balls as white. Suppose we
choose one of these bags at random. For this bag we select five balls at
random, replacing each ball after it has been selected. The result is that
we find 4 white balls and one black. What is the probability that we
were using the bag with mainly white balls?
Solution
Let A be the random variable "bag chosen" then A={a1,a2} where a1
represents "bag with mostly white balls" and a2 represents "bag with mostly
black balls" . We know that P(a1)=P(a2)=1/2 since we choose the bag at
random.
Let B be the event "4 white balls and one black ball chosen from 5
selections".
Then we have to calculate P(a1|B). From Bayes' rule this is:
Solution cont.
Now, for the bag with mostly white balls the probability of a ball being white
is ¾ and the probability of a ball being black is ¼. Thus, we can use the
Binomial Theorem, to compute P(B|a1) as:
Similarly
hence
Any Question???
Thank you

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Baye's rule

  • 1. Bayes' Rule Submitted By Md. Isarot Hossan ID: 142014001 Department of Textile Engineering Green University of Bangladesh
  • 2. Bayes' rule  True Bayesians actually consider conditional probabilities as more basic than joint probabilities. It is easy to define P(A|B) without reference to the joint probability P(A,B). To see this note that we can rearrange the conditional probability formula to get: P(A|B) P(B) = P(A,B)
  • 3. Bayes' Rule But by symmetry we can also get: P(B|A) P(A) = P(A,B) It follows that: Or, Which is the so-called Bayes' Rule.
  • 4. Bayes Rule Example  Suppose that we have two bags each containing black and white balls. One bag contains three times as many white balls as blacks. The other bag contains three times as many black balls as white. Suppose we choose one of these bags at random. For this bag we select five balls at random, replacing each ball after it has been selected. The result is that we find 4 white balls and one black. What is the probability that we were using the bag with mainly white balls?
  • 5. Solution Let A be the random variable "bag chosen" then A={a1,a2} where a1 represents "bag with mostly white balls" and a2 represents "bag with mostly black balls" . We know that P(a1)=P(a2)=1/2 since we choose the bag at random. Let B be the event "4 white balls and one black ball chosen from 5 selections". Then we have to calculate P(a1|B). From Bayes' rule this is:
  • 6. Solution cont. Now, for the bag with mostly white balls the probability of a ball being white is ¾ and the probability of a ball being black is ¼. Thus, we can use the Binomial Theorem, to compute P(B|a1) as: Similarly hence