# Example 16 - Chapter 13 Class 12 Probability

Last updated at April 16, 2024 by Teachoo

Examples

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Example 10

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Example 12 Important

Example 13 Important

Example 14 Important

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Example 16 You are here

Example 17 Important

Example 18 Important

Example 19 Important

Example 20 Important

Example 21 Important

Example 22 Important

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Example 24 Important

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

Serial order wise

Last updated at April 16, 2024 by Teachoo

Example 16 Bag I contains 3 red & 4 black balls while another Bag II contains 5 red & 6 black balls. One ball is drawn at random from one of bags & 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)) "P(E1)" = Probability bag selected is Bag I = 𝟏/𝟐 "P(E2)" = Probability bag selected is Bag II = 𝟏/𝟐 "P(A|E2)" = Probability red ball was selected from Bag II = 5/(5 + 6) = 𝟓/𝟏𝟏 "P(A|E1)" = Probability red ball was selected from Bag I = 3/(3 + 4) = 𝟑/𝟕 Putting values in formula, P("E2|"A) = (𝟏/𝟐 × 𝟓/𝟏𝟏)/( 𝟏/𝟐 × 𝟑/𝟕 + 𝟏/𝟐 × 𝟓/𝟏𝟏 ) = (1/2 × 5/11)/(1/2 [ 3/7 × 5/11 ]) = (5/11)/((33 + 35)/77) = (5/11)/( 68/77 ) = 5/11 × 77/68 = 𝟑𝟓/𝟔𝟖 Therefore, required probability is 𝟑𝟓/𝟔𝟖 = (1/2 × 5/11)/(1/2 [ 3/7 × 5/11 ]) = (5/11)/((33 + 35)/77) = (5/11)/( 68/77 ) = 5/11 × 77/68 = 𝟑𝟓/𝟔𝟖 Therefore, required probability is 𝟑𝟓/𝟔𝟖