In a box containing 15 bulbs 5 are defective
WebIn 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 Only one of them is defective Easy Solution … WebA 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 -
In a box containing 15 bulbs 5 are defective
Did you know?
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 % 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
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 … WebDoubtnut. In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is (A) 10-1 (B) ` (1/2)^5` (C) ` (9/ (10))^5` (D) `9/ (10)` …
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... WebFour light bulbs are chosen at random from 15 bulbs of which 5 are defective. Find the probability that none is defective. * 67/91 24/91 2/13 3/13 A box contains 8 spark plugs 1 of which is defective. If Robert picks 2 spark plugs from the box what is the probability that both spark plugs chosen are not defective? * 1/8 ¼ 2/8 ¾
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
WebWhat is the probability that out of a sample of 5 bulbs (i) none is defective (ii) exactly 2 are defective? Medium Solution Verified by Toppr Let p be the event of obtaining defective bulbs p= 606= 101,q=(1− 101)= 109,n=5 P(X=r)= nC r.p r.q (n−r)= 5C r(101)r.(109)5−r (i) P (none is defective) =P(X=0)= 5C 0(109)5=(109)5 birmingham rexelWebApr 7, 2024 · Now, let us find the probability to draw one non-defective bulbs and two defective bulbs. $ \Rightarrow P\left( X=2 \right)=\dfrac{\text{no}\text{. of ways of drawing one bulbs from 8 non-defective bulbs and two defective bulb from 5 defective bulbs}}{\text{no}\text{. of ways of drawing three bulbs from 13 bulbs in a box}} $ . birmingham rickwood fieldWebThe 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 dangerous machinery in the kitchenWebTranscribed 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. dangerous machinery signageWebNov 19, 2024 · Best answer. Out of 15 bulbs, 5 are defective probability of selecting a defective bulb = P = 5/15 = 1/3. We are selecting 5 bulbs n (S) = 15C5. (i) None of them is … dangerous magic tricksWebAnswer (1 of 2): There are 5 cases. 1. First bulb is defective. The probability is (5/15)(10/14)(9/13)(8/12)(7/11). 2. 2nd bulb is defective. The probability is (10/ ... birmingham rheumatology faxWebOct 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 birmingham rhinoplasty