Ex 13.3, 5 - Chapter 13 Class 12 Probability (Term 2)
Last updated at May 29, 2018 by Teachoo
Bayes theorem
Example 16
Ex 13.3, 2 Important
Ex 13.3, 3
Misc 3
Ex 13.3, 4 Important
Ex 13.3, 9
Ex 13.3, 5 You are here
Ex 13.3, 13 (MCQ) Important
Example 21 Important
Ex 13.3, 6 Important
Ex 13.3, 10 Important
Example 18 Important
Example 17 Important
Ex 13.3, 7
Ex 13.3, 8 Important
Example 19 Important
Ex 13.3, 11
Ex 13.3, 12 Important
Example 37 Important
Example 20 Important
Example 33 Important
Misc 12
Misc 16 Important
Misc 13 Important
Chapter 13 Class 12 Probability (Term 2)
Concept wise
- Conditional Probability - Values given
- Conditional Probability - Statement
- Multiplication rule of probability
- Independent events
- Basic Probability
- Theorem of total probability
-
Bayes theorem
- Random Variable - Defination
- Probability distribution
- Mean or Expectation of random variable
- Variance and Standard Deviation of a Random Variable
- Bernoulli Trials
- Binomial Distribution
Transcript
Ex 13.3, 5 A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e. if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive ? Let A : Person has the disease B : Person does not have the disease C : test result is positive We need to find the Probability that a person has the disease given that his test result is positive i.e. P(A|C) P(A|C) = P(A) . P(C|A) P(B) . P(C|B) + P(A) P(C|A) Putting values in formula, P(A|C) = 0. 001 0. 99 0. 999 0. 005 + 0. 001 0. 99 = 99 10 5 499. 5 10 5 + 0. 001 0.99 = 10 5 99 10 5 [ 499. 5 + 99 ] = 99 598. 5 = 990 5985 = Therefore, required probability is 22 133
