In a box containing 15 bulbs 5 are defective
WebA box contains 5 detective and 15 non-detective bulbs. Two bulbs are chosen at random. Find the probability that both the bulbs are non-defective. Answer: C) 21/38 Explanation: n (S) = C 2 20 = 190 n (E) = C 2 15 = 105 Therefore, P (E) = 105/190 = 21/38 Subject: Probability - Quantitative Aptitude - Arithmetic Ability Related Questions Q: WebQuestion: a 5. In a box containing 15 bulbs, 5 are defective. If 5 bulbs are selected at random from the box find the probability of the event, that i) none of them is defective ii) …
In a box containing 15 bulbs 5 are defective
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Web1 In the parentheses, you have the probability that both are defective or exactly one is defective. This is what you need to calculate. Don't subtract from $1$. (You could subtract the probability that both are good from $1$.) – David Mitra Feb 7, 2015 at 16:55 Add a comment 2 Answers Sorted by: 2 WebAnswer (1 of 2): There were total (35 + 15) = 50 bulbs in the box. If bulbs were drawn without replacement; probability of getting both defective bulbs out of two bulbs drawn = {(35C2) / (50C2)} = (595 / 1225) ≈ 0.4857 If bulbs were drawn with replacement; probability of getting both defective ...
WebSep 27, 2016 · (1) since fewer than 5 bulbs are defective we are limited to choices 4,3,2 or 1. So max 4 numbers to plug, not many. First lets take 4 defective bulbs. … WebTranscribed Image Text: Question 8 In a box containing 15 bulbs, 5 are defective. If 5 bulls are selected at random from the box find the probability of the event, that (i) none of them is defective (ii) only one of them is defective (iii) atleast one of them is defective.
WebThe material used for the bulbs in one of the two boxes was faulty so that one out of four bulbs go off as soon as you use them. The other box doesn't contain any fault bulbs. One box is selected at random and two bulbs are selected (without replacement) from it and tested. None of these two bulbs go off. WebAnswer (1 of 7): This is a nice problem which can be used to illustrate some basic principles. I’ll assume that the “at least one will be defective” means “at least one of the 15 bulbs removed will be defective - as otherwise the problem is trivial. First the Complement principle - Often it is e...
WebA box in a supply room contains 15 compact fluorescent lightbulbs, of which 5 are rated 13-watt, 6 are rated 18-watt, and 4 are rated 23-watt. Suppose that three of these bulbs are randomly selected. a. What is the probability that exactly two of …
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 … improper position on a highway orsWebOct 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 lithia motors earningsWebIn a box containing 15 bulbs, 5 are defective. If 5 bulbs are selected at random from the box, the probability of the event that (i) none of them is defectiv... improper platform usageWebSolution 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 improper plates vtl nyWebThe repeated selections of defective bulbs from a box are Bernoulli trials. Let X denote the number of defective bulbs out of a sample of 5 bulbs. Probability of getting a defective bulb, p= 10010 = 101 ∴q=1−p=1− 101 = 109 Clearly, X has a binomial distribution with n=5 and p= 101 ∴P(X=x)= nC xq n−xp x= 5C x(109)5−x(101)x improper placed fuel filterWebA box contains 5 radio tubes of which 2 are defective.The tubes are tested one after the other until the 2 defective tubes are discovered . Find the probability that the process stopped on the (i) Second test; (ii) third test, find the probability that the first tube is non-defective. Medium Solution Verified by Toppr Solution - improper punishment mlp fanfictionlithia motors/driveway finance