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Chapter 13 Class 12 Probability
Example 6
Ex 13.1, 10 (a) Important
Ex 13.1, 12 Important
Example 11 Important
Ex 13.2, 7 Important
Ex 13.2, 11 (i)
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Example 18 Important
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Example 21 Important
Ex 13.3, 2 Important
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Ex 13.3, 13 (MCQ) Important
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 5 Important Deleted for CBSE Board 2024 Exams
Question 6 Deleted for CBSE Board 2024 Exams
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Question 10 Important Deleted for CBSE Board 2024 Exams
Question 11 Important Deleted for CBSE Board 2024 Exams
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 6 Important Deleted for CBSE Board 2024 Exams
Question 10 Important Deleted for CBSE Board 2024 Exams
Question 13 Important Deleted for CBSE Board 2024 Exams
Question 13 Deleted for CBSE Board 2024 Exams
Example 23 Important
Question 2 Important Deleted for CBSE Board 2024 Exams
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Question 6 Important Deleted for CBSE Board 2024 Exams
Misc 7 Important
Misc 10 Important
Chapter 13 Class 12 Probability
Last updated at Aug. 16, 2023 by Teachoo
Example 7 Consider the experiment of tossing a coin. If the coin shows head, toss it again but if it shows tail, then throw a die. Find the conditional probability of the event that ‘the die shows a number greater than 4’ given that ‘there is at least one tail’.sLet’s Make a Tree Diagram We need to find the probability that the die shows a number greater than 4, given that there is at least one tail. Now, E: Number greater than 4 on the die F: at least one tail We need to find P(E|F) Event E E = { (T, 5), (T, 6) } P(E) = 1/12 + 1/12 = 𝟏/𝟔 Event F F = { (H, T), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6) } P(F) = 1/4 + 1/12 + 1/12 +1/12 +1/12 +1/12 + 1/12 + 1/12 = 1/4 + 6/12 = 1/4 + 1/2 = 𝟑/𝟒 Also, E ∩ F = {(T, 5), (T, 6)} P(E ∩ F) = 1/12 + 1/12 = 2/12 = 𝟏/𝟔 Now, P(E|F) = (𝑃(𝐸 ∩ 𝐹))/(𝑃(𝐹)) = (1/6)/(3/4) = 1/6 × 4/3 = 𝟐/𝟗