Question 6 - Chapter 13 Class 12 Probability (Important Question)
Last updated at April 16, 2024 by Teachoo
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
Ex 13.3, 12 Important
Ex 13.3, 13 (MCQ) Important
Question 4 Important Deleted for CBSE Board 2025 Exams
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Question 10 Important Deleted for CBSE Board 2025 Exams
Question 11 Important Deleted for CBSE Board 2025 Exams
Question 4 Important Deleted for CBSE Board 2025 Exams
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Question 10 Important Deleted for CBSE Board 2025 Exams
Question 13 Important Deleted for CBSE Board 2025 Exams
Question 13 Deleted for CBSE Board 2025 Exams
Example 23 Important
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Misc 7 Important
Misc 10 Important
Chapter 13 Class 12 Probability
Last updated at April 16, 2024 by Teachoo
Question 6 In a game, a man wins a rupee for a six and loses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amount he wins / loses.There will be 4 cases He gets 6 in first throw He does not get 6 in 1st throw , He gets six in 2nd throw He does not get 6 in 1st throw , He does not get 6 in 2nd throw, He gets six in 3rd throw He does not get 6 in 1st throw , 2nd throw or 3rd throw Case 1 He gets a six Probability = 1/6 Amount won = 1 Case 2 He does not get 6 in 1st throw , He gets six in 2nd throw So, P(win) = 5/6 × 1/6 = 5/36 Amount won = –1 + 1 = 0 Case 3 He does not get 6 in 1st throw , He does not get 6 in 2nd throw ,He gets six in 3rd throw So, P(win) = 5/6 × 5/6 × 1/6 = 25/216 Amount won = –1 – 1 + 1 = –1 Case 4 He does not get 6 in 1st throw , 2nd throw or 3rd throw So, P(win) = 5/6 × 5/6 × 5/6 = 125/216 Amount won = –1 – 1 − 1 = –3 Expected Value = Amount won × Probability for all four throws = (𝟏×𝟏/𝟔) + (𝟎×𝟓/𝟑𝟔) + (−𝟏×𝟐𝟓/𝟐𝟏𝟔) + (−𝟑×𝟏𝟐𝟓/𝟐𝟏𝟔) = 1/6 + 0 – 25/216 – 375/216 = 1/6 – 25/216 – 375/216 = (36 − 25 − 375)/216 = (36 − 400)/216 = (−364)/216 = (−𝟗𝟏)/𝟓𝟒