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 2024 Exams
Question 5 Important Deleted for CBSE Board 2024 Exams
Question 6 Deleted for CBSE Board 2024 Exams You are here
Question 7 Important Deleted for CBSE Board 2024 Exams
Question 8 Important Deleted for CBSE Board 2024 Exams
Question 3 Important Deleted for CBSE Board 2024 Exams
Question 6 Important Deleted for CBSE Board 2024 Exams You are here
Question 11 Important Deleted for CBSE Board 2024 Exams
Question 15 Deleted for CBSE Board 2024 Exams
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 You are here
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 You are here
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