# Misc 10 - Chapter 13 Class 12 Probability (Term 2)

Last updated at Feb. 15, 2020 by Teachoo

Miscellaneous

Misc 1 (i)

Misc 1 (ii)

Misc 2 (i) Important

Misc 2 (ii)

Misc 3

Misc 4 Deleted for CBSE Board 2022 Exams

Misc 5 Important Deleted for CBSE Board 2022 Exams

Misc 6 Important Deleted for CBSE Board 2022 Exams

Misc 7 Important

Misc 8 Important

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Misc 10 Important You are here

Misc 11 Important

Misc 12

Misc 13 Important

Misc 14 Important

Misc 15 Important

Misc 16 Important

Misc 17 (MCQ) Important

Misc 18 (MCQ)

Misc 19 (MCQ)

Chapter 13 Class 12 Probability (Term 2)

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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 𝒒^(𝒏−𝒙) 𝒑^𝒙 Here, n = number of coins tosses p = Probability of head = 1/2 q = 1 – p = 1 – 1/2 = 1/2 Hence, P(X = x) = nCx (1/2)^𝑥 (1/2)^(𝑛−𝑥) P(X = x) = nCx (1/2)^(𝑛 − 𝑥 + 𝑥) P(X = x) = nCx (𝟏/𝟐)^𝒏 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 ≥ 1) > 90%, we need to find n Now, P(X ≥ 1) > 90 % 1 − P(X = 0) > 90 % 1 − P(X = 0) > 90 % 1 − nC0 (1/2)^𝑛> 90 % 1 − 1/2^𝑛 > 90/100 1 − 1/2^𝑛 > 9/10 1 − 9/10 > 1/2^𝑛 (10 − 9)/10 > 1/2^𝑛 1/10 > 1/2^𝑛 𝟐^𝒏 > 10 We know that 24 = 16 > 10 So, n ≥ 4.