# Question 13 - Examples - Chapter 13 Class 12 Probability

Last updated at April 16, 2024 by Teachoo

Examples

Example 1

Example 2

Example 3

Example 4

Example 5 Important

Example 6

Example 7 Important

Example 8

Example 9 Important

Example 10

Example 11 Important

Example 12 Important

Example 13 Important

Example 14 Important

Example 15 Important

Example 16

Example 17 Important

Example 18 Important

Example 19 Important

Example 20 Important

Example 21 Important

Example 22 Important

Example 23 Important

Example 24 Important

Question 1 Deleted for CBSE Board 2025 Exams

Question 2 Deleted for CBSE Board 2025 Exams

Question 3 Important Deleted for CBSE Board 2025 Exams

Question 4 Important Deleted for CBSE Board 2025 Exams

Question 5 Important Deleted for CBSE Board 2025 Exams

Question 6 Deleted for CBSE Board 2025 Exams

Question 7 Important Deleted for CBSE Board 2025 Exams

Question 8 Important Deleted for CBSE Board 2025 Exams

Question 9 Important Deleted for CBSE Board 2025 Exams

Question 10 Important Deleted for CBSE Board 2025 Exams

Question 11 Important Deleted for CBSE Board 2025 Exams

Question 12 Deleted for CBSE Board 2025 Exams

Question 13 Deleted for CBSE Board 2025 Exams You are here

Chapter 13 Class 12 Probability

Serial order wise

Last updated at April 16, 2024 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 `