WebThree light bulbs are chosen at random from 15 bulbs of which 5 are defective. What is he probability that atleast one is defective? Question Transcribed Image Text: Three light bulbs are chosen at random from 15 bulbs of which 5 are defectíve. What is ihe probability that atleast one is defective? Expert Solution Want to see the full answer? Webp = probability of getting a defective bulb = 5 15 = 1 3 and q = probability of getting non-defective bulb = 1-p = 1-1 3 = 2 3 Let X denote the number of success in a sample of 4 trials. ... A box contains 13 bulbs, out of which 5 are defective. 3 bulbs are randomly drawn, one by one without replacement, from the box. Find the probability ...
Ex 15.1, 17 - A lot of 20 bulbs contain 4 defective ones - teachoo
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. WebA box of 15 flashbulbs contains 5 defective bulbs. A random sample of 2 is selected and tested. Let X be the random variable associated with the number of defective bulbs in the … daughtry\u0027s wife
A box contains 10 items, 3 of which are defective. If 4 are ... - Toppr
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) … 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 WebEasy Solution Verified by Toppr Correct option is D) Here at least 2 items are defective out of 4. So the no. of defective items can be 2 or defective can be 3. Total no. of defective ites =3 If 2 items are defective = 3C 2× 7C 2=63 If 3 items are defective = 3C 3× 7C 1=7 black 270s size 5