WebJun 29, 2024 · Number of light bulbs in the box = 20. Number of defective light bulbs= 5. So, non defective light bulb= 20-5=15. Probability of an event . Now, 4 light bulbs are picked randomly,the probability that at most 2 of them are defective is =0.9680. Required probability = 0.97 or 97 % WebIf a person selects $4$ bulbs from the box at random, without replacement, what is the probability that all $4$ bulbs will be defective? solution: there are $4!$ ways to select the …
From a lot of 10 bulbs, which includes 3 defectives bulbs, a
WebJan 29, 2009 · A box contains 24 light bulbs of which 4 are defective. If one person selects 10 bulbs from the box in a random manner, and a second person then takes the remaining 14 bulbs, what is the probability that all 4 detective bulbs will be obtained by the same person.? Homework Equations WebDec 27, 2024 · In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, Doubtnut 2.49M subscribers Subscribe 8 Share 665 views 2 years ago In a box... trace insight security
How to find the probability of a defective bulb?
WebIn a box containing 100 bulbs, 10 are defective. What\\( \\mathrm{P} \\) is the probability that out of a sample of 5 bulbs, noneW is defective?(1) \\( 10^{-5} \\... WebOct 8, 2024 · What is the probability that among 5 bulbs chosen at random, none is defective? probability class-11 1 Answer +1 vote answered Oct 8, 2024 by Anjali01 (48.1k points) selected Oct 8, 2024 by RamanKumar Best answer Total number of bulbs = 10 Number of defective bulbs = 2 ∴ Number of good bulbs = 10 – 2 = 8 WebSolution Verified by Toppr There are 3 defective bulbs and 7 non-defective bulbs. Let x denote the random vanable of the no.of defective bulb. Then x can take values 0,1,2 since bulbs are replaced. p=p(D) 10$$3 q=p(D)=1− 103 = 107 p(x=0)= 10c 27c 2×3c 0 = 10×47×6 = 157 p (x=1)= 10 27 13c 2 = 10×91×3×2 = 157 p(x=2)= 10 127c 0×3c 2 = 10×91×3×2 = 151 trace insight education