# Example 22 - Chapter 13 Class 12 Probability

Last updated at April 16, 2024 by Teachoo

Examples

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Example 22 Important You are here

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Question 1 Deleted for CBSE Board 2025 Exams

Question 2 Deleted for CBSE Board 2025 Exams

Question 3 Important Deleted for CBSE Board 2025 Exams

Question 4 Important Deleted for CBSE Board 2025 Exams

Question 5 Important Deleted for CBSE Board 2025 Exams

Question 6 Deleted for CBSE Board 2025 Exams

Question 7 Important Deleted for CBSE Board 2025 Exams

Question 8 Important Deleted for CBSE Board 2025 Exams

Question 9 Important Deleted for CBSE Board 2025 Exams

Question 10 Important Deleted for CBSE Board 2025 Exams

Question 11 Important Deleted for CBSE Board 2025 Exams

Question 12 Deleted for CBSE Board 2025 Exams

Question 13 Deleted for CBSE Board 2025 Exams

Chapter 13 Class 12 Probability

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Last updated at April 16, 2024 by Teachoo

Example 22 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)= (๐(๐ธ3).๐(๐ด|๐ธ3))/(๐(๐ธ1)๐(๐ด|๐ธ1)+๐(๐ธ2)๐(๐ด|๐ธ2)+๐(๐ธ3)๐(๐ด|๐ธ3)+๐(๐ธ4)๐(๐ด|๐ธ4)) P(E1) = Probability that ball drawn is from box I = ๐/๐ P(A|E1) = Probability of that black ball is selected from Box I = (๐๐ข๐๐๐๐ ๐๐ ๐๐๐๐๐ )/(๐๐๐ก๐๐ ๐๐ข๐๐๐๐ ๐๐ ๐๐๐๐๐ ๐๐ ๐กโ๐ ๐๐๐ฅ) = ๐/๐๐ P(E2) = Probability that ball drawn is from box II = ๐/๐ P(A|E2) = Probability of that black ball is selected from Box II = (๐๐ข๐๐๐๐ ๐๐ ๐๐๐๐๐ )/(๐๐๐ก๐๐ ๐๐ข๐๐๐๐ ๐๐ ๐๐๐๐๐ ๐๐ ๐กโ๐ ๐๐๐ฅ) = 2/8 = 1/4 P(E3) = Probability that ball drawn is from box III = ๐/๐ P(A|E3) = Probability of that black ball is selected from Box II = (๐๐ข๐๐๐๐ ๐๐ ๐๐๐๐๐ )/(๐๐๐ก๐๐ ๐๐ข๐๐๐๐ ๐๐ ๐๐๐๐๐ ๐๐ ๐กโ๐ ๐๐๐ฅ) = ๐/๐ P(E4) = Probability that ball drawn is from box IV = ๐/๐ P(A|E4) = Probability of that black ball is selected from Box IV = (๐๐ข๐๐๐๐ ๐๐ ๐๐๐๐๐ )/(๐๐๐ก๐๐ ๐๐ข๐๐๐๐ ๐๐ ๐๐๐๐๐ ๐๐ ๐กโ๐ ๐๐๐ฅ) = ๐/๐๐ Putting values in Equation "P(E3|A)"=(๐(๐ธ3).๐(๐ด|๐ธ3))/(๐(๐ธ1)๐(๐ด|๐ธ1)+๐(๐ธ2)๐(๐ด|๐ธ2)+๐(๐ธ3)๐(๐ด|๐ธ3)+๐(๐ธ4)๐(๐ด|๐ธ4)) = (๐/๐ ร ๐/๐)/( ๐/๐ ร ๐/๐๐ + ๐/๐ ร ๐/๐ + ๐/๐ ร ๐/๐ + ๐/๐ ร ๐/๐๐ ) = (1/28)/( 1/24 + 1/16 + 1/28 + 1/13 ) = (1/28)/( (1/(4 ร 6) + 1/(4 ร 4) + 1/(4 ร 7)) + 1/13) = (1/28)/( ((4 ร 7 + 6 ร 7 + 6 ร 4)/(4 ร 6 ร 4 ร 7)) + 1/13) = (1/28)/((28 + 42 + 24)/(4 ร 6 ร 4 ร 7) + 1/13) = (1/28)/(94/(4 ร 6 ร 4 ร 7) + 1/13) = (1/28)/((94 ร 13 + 4 ร 6 ร 4 ร 7)/(13 ร 4 ร 6 ร 4 ร 7) ) = 1/((94 ร 13 + 4 ร 6 ร 4 ร 7)/(13 ร 4 ร 6) ) = (13 ร 4 ร 6)/( 94 ร 13 + 4 ร 6 ร 4 ร 7) = (13 ร 2 ร 6)/( 47 ร 13 + 2 ร 6 ร 4 ร 7) = 156/(611 + 336) = 156/947 = 0.164