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Chapter 13 Class 12 Probability
Example 6
Ex 13.1, 10 (a) Important
Ex 13.1, 12 Important
Example 11 Important
Ex 13.2, 7 Important
Ex 13.2, 11 (i)
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Example 18 Important
Example 20 Important
Example 21 Important
Ex 13.3, 2 Important
Ex 13.3, 4 Important
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Ex 13.3, 10 Important
Ex 13.3, 12 Important
Ex 13.3, 13 (MCQ) Important
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 5 Important Deleted for CBSE Board 2024 Exams
Question 6 Deleted for CBSE Board 2024 Exams
Question 7 Important Deleted for CBSE Board 2024 Exams
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Question 3 Important Deleted for CBSE Board 2024 Exams
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Question 11 Important Deleted for CBSE Board 2024 Exams
Question 15 Deleted for CBSE Board 2024 Exams
Question 10 Important Deleted for CBSE Board 2024 Exams
Question 11 Important Deleted for CBSE Board 2024 Exams
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 6 Important Deleted for CBSE Board 2024 Exams
Question 10 Important Deleted for CBSE Board 2024 Exams
Question 13 Important Deleted for CBSE Board 2024 Exams You are here
Question 13 Deleted for CBSE Board 2024 Exams You are here
Example 23 Important
Question 2 Important Deleted for CBSE Board 2024 Exams
Question 4 Deleted for CBSE Board 2024 Exams
Question 6 Important Deleted for CBSE Board 2024 Exams
Misc 7 Important
Misc 10 Important
Chapter 13 Class 12 Probability
Last updated at May 29, 2023 by Teachoo
Question 13 The probability of a shooter hitting a target is 3/4 . How many minimum number of times must he/she fire so that the probability of hitting the target at least once is more than 0.99?Let X : Number of times he hits the target Hitting the target is a Bernoulli trial So, X has a binomial distribution P(X = x) = nCx 𝒒^(𝒏−𝒙) 𝒑^𝒙 Here, n = number of rounds fired p = Probability of hitting = 3/4 q = 1 – p = 1 − 3/4 = 1/4 Hence, P(X = x) = nCx (𝟑/𝟒)^𝒙 (𝟏/𝟒)^(𝒏−𝒙) We need to find How many minimum number of times must he/she fire so that the probability of hitting the target at least once is more than 0.99 So, given P(X ≥ 1) > 99%, we need to find n Now, P(X ≥ 1) > 99 % 1 − P(X = 0) > 99 % ` 1 − nC0(3/4)^0 (1/4)^𝑛> 0.99 1 − (1/4)^𝑛 > 0.99 1 − 0.99 > (1/4)^𝑛 0.01 > 1/4^𝑛 4^𝑛 > 1/0.01 𝟒^𝒏 > 𝟏𝟎𝟎 We know that 44 = 256 So, n ≥ 4 So, the minimum value of n is 4 So, he must fire atleast 4 times `