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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
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Example 21 Important
Ex 13.3, 2 Important
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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
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Question 8 Important Deleted for CBSE Board 2024 Exams
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 You are here
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 You are here
Question 13 Important Deleted for CBSE Board 2024 Exams
Question 13 Deleted for CBSE Board 2024 Exams
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 10 If a fair coin is tossed 10 times, find the probability of (i) exactly six heads (ii) at least six heads (iii) at most six headsIf a trial is Bernoulli, then There is finite number of trials They are independent Trial has 2 outcomes i.e. Probability success = P then Probability failure = q = 1 – P (4) Probability of success (P) is same for all trials 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 n = number of coins tosses = 10 p = Probability of head = 1/2 q = 1 – p = 1 – 1/2 = 1/2 Hence, P(X = x) = 10Cx (1/2)^𝑥 (1/2)^(10−𝑥) P(X = x) = 10Cx (1/2)^(10 − 𝑥 + 𝑥) P(X = x) = 10Cx (𝟏/𝟐)^𝟏𝟎 Probability exactly six heads Probability exactly six heads = P(X = 6) Putting x = 6 in (1) P(X = 6) = 10C6 (1/2)^10 = (10 !)/((10 − 6) ! ×6 !) × (1/2)^10= (10 !)/(4 ! × 6 !) × 1/2^10 = 105/512 (ii) Probability appearing at least six heads i.e. P(X ≥ 6) P(X ≥ 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 10C6 (1/2)^10 + 10C7 (1/2)^10 + 10C8 (1/2)^10 + 10C9 (1/2)^10 + 10C10 (1/2)^10 = (1/2)^10(10C6 + 10C7 + 10C8 + 10C9 + 10C10) = (1/2)^10(210 + 120 + 45 + 10 + 1) = (1/2)^10(386) = 𝟏𝟗𝟑/𝟓𝟏𝟐 (iii) Probability appearing at most six heads i.e. P(X ≤ 6) P(X ≤ 6) = P(X = 6) + P(X = 5) + P(X = 4) + P(X = 3) + P(X = 2) + P(X = 1) + P(X = 0) = 10C6 (1/2)^10 + 10C5 (1/2)^10 + 10C4 (1/2)^10 + 10C3 (1/2)^10 + 10C2 (1/2)^10 + 10C1 (1/2)^10+ 10C0 (1/2)^10 = (1/2)^10(10C6 + 10C5 + 10C4 + 10C3 + 10C2 + 10C1 + 10C0) = (1/2)^10(210 + 252 + 210 + 120 + 45 + 10 + 1) = (1/2)^10(848) = 𝟓𝟑/𝟔𝟒