


Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Miscellaneous
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 2024 Exams
Question 2 Important Deleted for CBSE Board 2024 Exams
Question 3 Important Deleted for CBSE Board 2024 Exams You are here
Question 4 Deleted for CBSE Board 2024 Exams
Question 5 Important Deleted for CBSE Board 2024 Exams
Question 6 Important Deleted for CBSE Board 2024 Exams
Last updated at May 29, 2023 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