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Example 16 - Chapter 13 Class 12 Probability (Term 2)
Last updated at May 29, 2018 by Teachoo
Bayes theorem
Example 16
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Ex 13.3, 2 Important
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Misc 3
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Ex 13.3, 13 (MCQ) Important
Example 21 Important
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Ex 13.3, 10 Important
Example 18 Important
Example 17 Important
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Example 19 Important
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Example 37 Important
Example 20 Important
Example 33 Important
Misc 12
Misc 16 Important
Misc 13 Important
Chapter 13 Class 12 Probability
Concept wise
- Conditional Probability - Values given
- Conditional Probability - Statement
- Multiplication rule of probability
- Independent events
- Basic Probability
- Theorem of total probability
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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
Example 16 Bag I contains 3 red and 4 black balls while another Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from Bag II. Let E1 : Bag selected is bag 1 E2 : Bag selected is bag 2 A : Ball selected is Red B : Ball selected is Black We need to find P(ball was drawn from bag 2, if ball is red) = P(E2|A) We need to find, 𝑃 𝐸2|𝐴= 𝑃 𝐴 𝑃(𝐴|𝐸2)𝑃 𝐸1 𝑃 𝐴|𝐸1 + 𝑃 𝐸2 𝑃(𝐴|𝐸2) Putting values in formula, P(E2|A) = 12 × 511 12 × 37 + 12 × 511 = 12 × 511 12 [ 37 × 511 ] = 511 33 + 3577 = 511 6877 = 511 × 7768 = 3568 Therefore, required probability is 𝟑𝟓𝟔𝟖
