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  1. Chapter 13 Class 12 Probability
  2. Serial order wise

Transcript

Misc 10 How many times must a man toss a fair coin so that the probability of having at least one head is more than 90%? Let X : Number of heads appearing Coin toss is a Bernoulli trial So, X has a binomial distribution P(X = x) = nCx ๐’’^(๐’โˆ’๐’™) ๐’‘^๐’™ Here, n = number of coins tosses p = Probability of head = 1/2 q = 1 โ€“ p = 1 โ€“ 1/2 = 1/2 Hence, P(X = x) = nCx (1/2)^๐‘ฅ (1/2)^(๐‘›โˆ’๐‘ฅ) P(X = x) = nCx (1/2)^(๐‘› โˆ’ ๐‘ฅ + ๐‘ฅ) P(X = x) = nCx (๐Ÿ/๐Ÿ)^๐’ We need to find How many times must a man toss a fair coin so that the probability of having at least one head is more than 90%? So, given P(X โ‰ฅ 1) > 90%, we need to find n Now, P(X โ‰ฅ 1) > 90 % 1 โˆ’ P(X = 0) > 90 % 1 โˆ’ P(X = 0) > 90 % 1 โˆ’ nC0 (1/2)^๐‘›> 90 % 1 โˆ’ 1/2^๐‘› > 90/100 1 โˆ’ 1/2^๐‘› > 9/10 1 โˆ’ 9/10 > 1/2^๐‘› (10 โˆ’ 9)/10 > 1/2^๐‘› 1/10 > 1/2^๐‘› ๐Ÿ^๐’ > 10 We know that 24 = 16 > 10 So, n โ‰ฅ 4.

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.