# Question 3 - Miscellaneous - Chapter 13 Class 12 Probability

Last updated at April 16, 2024 by Teachoo

Miscellaneous

Misc 1 (i)

Misc 1 (ii)

Misc 2 (i) Important

Misc 2 (ii)

Misc 3

Misc 4

Misc 5 Important

Misc 6

Misc 7 Important

Misc 8 Important

Misc 9 Important

Misc 10 Important

Misc 11 (MCQ) Important

Misc 12 (MCQ)

Misc 13 (MCQ)

Question 1 Important Deleted for CBSE Board 2025 Exams

Question 2 Important Deleted for CBSE Board 2025 Exams

Question 3 Important Deleted for CBSE Board 2025 Exams You are here

Question 4 Deleted for CBSE Board 2025 Exams

Question 5 Important Deleted for CBSE Board 2025 Exams

Question 6 Important Deleted for CBSE Board 2025 Exams

Chapter 13 Class 12 Probability

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Last updated at April 16, 2024 by Teachoo

Question 3 A die is thrown again and again until three sixes are obtained. Find the probability of obtaining the third six in the sixth throw of the die. We need to find probability of obtaining the third six in the sixth throw of the die. P(getting 3rd six in 6th throw) = P(getting 2 sixes in 5 throws) × P(getting a six on 6th throw) = P(getting 2 sixes in 5 throws) × 1/6 Calculating P(getting 2 sixes in 5 throws) Let X : be the number six we get on 5 throws Throwing a pair of die is a Bernoulli trial So, X has binomial distribution P(X = x) = nCx 𝒒^(𝒏−𝒙) 𝒑^𝒙 Here n = number of times die is thrown = 5 p = Probability of getting a six = 1/6 q = 1 – 1/6 = 5/6 Hence, P(X = x) = 5Cx (𝟏/𝟔)^𝒙 (𝟓/𝟔)^(𝟓 − 𝒙) We need to find P(getting 2 sixes in 5 throws) i.e. P(X = 2) P(X = 2) = 5C2 (1/6)^2 (5/6)^(5 − 2 ) = 5!/( (5 − 2) ! 2 !) (1/6)^2 (5/6)^3 = (5 × 4 × 3! )/(3 ! × 2 !) 1/(6 × 6) × (5/6)^3 = 10/(6^2 ) (5/6)^3 = (10 ×〖 5〗^3)/(6^2 × 6^3 ) =(10 ×〖 5〗^3)/6^5 Hence, Required Probability = P(X = 2) × 1/6 = (10 ×〖 5〗^3)/6^5 ×1/6 = (10 ×〖 5〗^3)/6^6 = (10 × 125)/46656 = 𝟔𝟐𝟓/𝟐𝟑𝟑𝟐𝟖