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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)
Ex 13.2, 14 Important
Example 17 Important
Example 18 Important
Example 20 Important
Example 21 Important
Ex 13.3, 2 Important
Ex 13.3, 4 Important
Ex 13.3, 8 Important
Ex 13.3, 10 Important You are here
Ex 13.3, 12 Important
Ex 13.3, 13 (MCQ) Important
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 5 Important Deleted for CBSE Board 2024 Exams
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Question 11 Important 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
Question 4 Deleted for CBSE Board 2024 Exams
Question 6 Important Deleted for CBSE Board 2024 Exams
Misc 7 Important
Misc 10 Important
Chapter 13 Class 12 Probability
Last updated at May 29, 2023 by Teachoo
Ex 13.3, 10 Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?Let A : 1, 2, 3, 4 appear on the die B : 5, 6 appear on the die C : exactly one head is obtained We need to find the Probability that if exactly one head is obtained on the toss of a coin, she threw 1, 2, 3 or 4 with the die i.e. P(A|C) P(A|C) = (𝑃(𝐴) ." " 𝑃(𝐶|𝐴))/(𝑃(𝐴) . 𝑃(𝐶|𝐴) + 𝑃(𝐵) . 𝑃(𝐶|𝐵) ) "P(A)" = Probability that 1, 2, 3 or 4 appear on the die = 4/6 = 𝟐/𝟑 "P(C|A)" = Probability that exactly one head is obtained, if 1, 2, 3 or 4 appear on the die = 𝟏/𝟐 "P(B)" = Probability that 5 or 6 appear on the die = 2/6 = 𝟏/𝟑 "P(C|B)" = Probability that exactly one head is obtained, if 5 or 6 appear on the die = 𝟑/𝟖 Puttinag values in formula, P(A"|"C) = (2/3 × 1/2)/( 2/3 × 1/2 + 1/3 × 3/8 ) = (1/3 × 1/2 × 2)/( 1/3 × 1/2 [2 + 3/4 ] ) = 2/( 2 + 3/4 ) = 2/(11/4) = 𝟖/𝟏𝟏 Therefore, required probability is 8/11