


Binomial Distribution
Ex 13.5, 12 Deleted for CBSE Board 2022 Exams
Ex 13.5, 2 Deleted for CBSE Board 2022 Exams
Ex 13.5, 4 Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 9 Deleted for CBSE Board 2022 Exams
Ex 13.5, 6 Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 11 Deleted for CBSE Board 2022 Exams
Ex 13.5, 14 (MCQ) Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 15 (MCQ) Important Deleted for CBSE Board 2022 Exams
Example 32 Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 13 Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 3 Important Deleted for CBSE Board 2022 Exams
Misc 6 Important Deleted for CBSE Board 2022 Exams
Misc 7 Important
Ex 13.5, 7 Important Deleted for CBSE Board 2022 Exams
Example 31 Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 5 Deleted for CBSE Board 2022 Exams You are here
Ex 13.5, 10 Important Deleted for CBSE Board 2022 Exams
Misc 5 Important Deleted for CBSE Board 2022 Exams
Misc 9 Deleted for CBSE Board 2022 Exams
Misc 4 Deleted for CBSE Board 2022 Exams
Misc 10 Important
Example 35 Deleted for CBSE Board 2022 Exams
Ex 13.5, 8 Deleted for CBSE Board 2022 Exams
Example 34 Deleted for CBSE Board 2022 Exams
Binomial Distribution
Ex 13.5, 5 The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs (i) none (ii) not more than one (iii) more than one (iv) at least one will fuse after 150 days of use.Let X : Number of bulbs fused Picking a bulb is a Bernoulli trial So, X has a binomial distribution P(X = x) = nCx 𝒒^(𝒏−𝒙) 𝒑^𝒙 Here n = number of bulbs picked = 5 p = Probability of getting fused bulb = 0.05 q = 1 − p = 1 − 0.05 = 0.95 Hence, P(X = x) = 5Cx (𝟎.𝟎𝟓)^𝒙 (𝟎.𝟗𝟓)^(𝟓−𝒙) Probability that out of 5 such bulbs none fuses i.e. P(X = 0) P(X = 0) = 5C0 (0.05)^0 (0.95)^(5−0) = 1 × 1 × (0.95)5 = (0.95)5 (ii) Probability that out of 5 such bulbs not more than one fuses P(not not than one) = P( X ≤ 1) = P(X = 0) + P(X = 1) = 5C0 (0.05)^0 (0.95)^(5−0) + 5C1 (0.05)^1 (0.95)^(5−1) = 1 × 1 × (0.95)5 + 5 × 0.05 × (0.95)4 = (0.95)4 [0.95+0.25] = (0.95)4 × 1.2 (iii) Probability that out of 5 such bulbs more than one fuses P (more than one) = 1 − P (not more than one) = 1 − ((0.95)4 × 1.2) (iv) probability that out of 5 such bulbs atleast one fuses P (At least one) = 1 − P (not even one fuses ) = 1 − P (0) = 1 − (0.95)5