Example 33 - Chapter 13 Class 12 Probability (Term 2)
Last updated at Feb. 15, 2020 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
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 You are here
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
Example 33 Coloured balls are distributed in four boxes as shown in the following table: A box is selected at random and then a ball is randomly drawn from the selected box. The colour of the ball is black, what is the probability that ball drawn is from the boxs III?Let A : Event that a black ball is selected E1 : Event that the ball is selected from box I E2 : Event that the ball is selected from box II E3 : Event that the ball is selected from box III E4 : Event that the ball is selected from box IV We need to find out the Probability of the ball drawn is from box III if it is black. i.e. P(E3|A) P(E3|A)= (π(πΈ3).π(π΄|πΈ3))/(π(πΈ1)π(π΄|πΈ1)+π(πΈ2)π(π΄|πΈ2)+π(πΈ3)π(π΄|πΈ3)+π(πΈ4)π(π΄|πΈ4)) P(E1) = Probability that ball drawn is from box I = 1/4 P(A|E1) = Probability of that black ball is selected from Box I = (ππ’ππππ ππ πππππ )/(πππ‘ππ ππ’ππππ ππ πππππ ππ π‘βπ πππ₯) = 3/18 P(E2) = Probability that ball drawn is from box II = 1/4 P(A|E2) = Probability of that black ball is selected from Box II = (ππ’ππππ ππ πππππ )/(πππ‘ππ ππ’ππππ ππ πππππ ππ π‘βπ πππ₯) = 2/8 = 1/4 P(E3) = Probability that ball drawn is from box III = 1/4 P(A|E3) = Probability of that black ball is selected froAm Box II = (ππ’ππππ ππ πππππ )/(πππ‘ππ ππ’ππππ ππ πππππ ππ π‘βπ πππ₯) = 1/7 P(E4) = Probability that ball drawn is from box IV = 1/4 P(A|E4) = Probability of that black ball is selected from Box IV = (ππ’ππππ ππ πππππ )/(πππ‘ππ ππ’ππππ ππ πππππ ππ π‘βπ πππ₯) = 4/13 Putting values in Equation "P(E3|A)"=(π(πΈ3).π(π΄|πΈ3))/(π(πΈ1)π(π΄|πΈ1)+π(πΈ2)π(π΄|πΈ2)+π(πΈ3)π(π΄|πΈ3)+π(πΈ4)π(π΄|πΈ4)) = (1/4 Γ 1/7)/( 1/4 Γ 3/18 + 1/4 Γ 1/4 + 1/4 Γ 1/7 + 1/4 Γ 4/13 ) = (1/28)/( 1/24 + 1/16 + 1/28 + 1/13 ) = (1/28)/( (1/(4 Γ 6) + 1/(4 Γ 4) + 1/(4 Γ 7)) + 1/13) = (1/28)/( ((4 Γ 7 + 6 Γ 7 + 6 Γ 4)/(4 Γ 6 Γ 4 Γ 7)) + 1/13) = (1/28)/((28 + 42 + 24)/(4 Γ 6 Γ 4 Γ 7) + 1/13) = (1/28)/(94/(4 Γ 6 Γ 4 Γ 7) + 1/13) = (1/28)/((94 Γ 13 + 4 Γ 6 Γ 4 Γ 7)/(13 Γ 4 Γ 6 Γ 4 Γ 7) ) = 1/((94 Γ 13 + 4 Γ 6 Γ 4 Γ 7)/(13 Γ 4 Γ 6) ) = (13 Γ 4 Γ 6)/( 94 Γ 13 + 4 Γ 6 Γ 4 Γ 7) = (13 Γ 2 Γ 6)/( 47 Γ 13 + 2 Γ 6 Γ 4 Γ 7) = 156/(611 + 336) = 156/947 = 0.164
