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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Question 1 A die is thrown 6 times. If β€˜getting an odd number’ is a success, what is the probability of (i) 5 successes? (ii) at least 5 successes? (iii) at most 5 successes?Let X : Number of times we get odd numbers in 6 throws of die Throwing a die is a Bernoulli trial So, X has a binomial distribution P(X = x) = nCx 𝒒^(π’βˆ’π’™) 𝒑^𝒙 Here, n = number of throws = 6 p = Probability of getting an odd number = 3/6 = 1/2 q = 1 – p = 1 – 1/2 = 1/2 Hence, P(X = x) = 6Cx (1/2)^π‘₯ (1/2)^(6βˆ’π‘₯) = 6Cx (1/2)^(6 βˆ’ π‘₯ + π‘₯) = 6Cx (𝟏/𝟐)^πŸ” Probability 5 successes Probability 5 success = P(X = 5) Putting x = 5 in (1) P(X = 5) = 6C5 (1/2)^6 = 6 Γ— 1/64 = πŸ‘/πŸ‘πŸ (ii) Probability appearing at least 5 successes i.e. P(X β‰₯ 5) P(X β‰₯ 5) = P(X = 5) + P(X = 6) = 6C5 (1/2)^6+ 6C6 (1/2)^6 = 6 (1/2)^6 + (1/2)^6 = 7(1/2)^6 = πŸ•/πŸ”πŸ’ (iii) Probability appearing at most 5 success i.e. P(X ≀ 5) P(X ≀ 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 6C0 (1/2)^6 + 6C1 (1/2)^6 + 6C2 (1/2)^6 + 6C3 (1/2)^6 + 6C4 (1/2)^6+ 6C5 (1/2)^6 = (1/2)^6 (6C0 + 6C1 + 6C2 + 6C3 + 6C4 + 6C5) = (1/2)^10(1 + 6 + 15 + 20 + 15 + 6) = πŸ”πŸ‘/πŸ”πŸ’

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.