Check sibling questions

Chapter 13 Class 12 Probability

Ex 13.4, 6 - From a lot of 30 bulbs which include 6 defectives, a samp

Ex 13.4, 6 - Chapter 13 Class 12 Probability - Part 2
Ex 13.4, 6 - Chapter 13 Class 12 Probability - Part 3

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Question 6 From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.Let X : be the number of defective bulbs Picking bulbs is a Bernoulli trial So, X has binomial distribution P(X = x) = nCx 𝒒^(π’βˆ’π’™) 𝒑^𝒙 n = number of times we pick a bulb = 4 p = Probability of getting defective bulb = 6/30 = 1/5 q = 1 – p = 1 – 1/5 = 4/5 Hence, P(X = x) = 4Cx (𝟏/πŸ“)^𝒙 (πŸ’/πŸ“)^(πŸ’βˆ’π’™) Now, P(X = 0) = 4C0 (1/5)^0 (4/5)^(4βˆ’0)= 4C0 (1/5)^0 (4/5)^4 = 256/625 P(X = 1) = 4C1 (1/5)^1 (4/5)^(4βˆ’1)= 4C1 (1/5)^1 (4/5)^3 = 4 Γ— 64/625 = 256/625 P(X = 2) = 4C2 (1/5)^2 (4/5)^(4βˆ’2)= 4C2 (1/5)^2 (4/5)^2 = 6 Γ— 16/625 = 96/625 P(X = 3) = 4C3 (1/5)^3 (4/5)^(4βˆ’3)= 4C3 (1/5)^3 (4/5)^1 = 4 Γ— 4/625 = 16/625 P(X = 4) = 4C4 (1/5)^4 (4/5)^(4βˆ’4)= 4C4 (1/5)^4 (4/5)^0= 1/625 So, the probability distribution is

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.