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om Student/PlayerHomework aspxhomeworklid 4640955088questionld-3aflushed false&icld-4828602; 8center BUS 352 TR700 Spring 2018 michael mccabe | 2/6/18 9:03 AM Homework Homework: Chapter 5 core: 0.25 of 1 pt 3016(5complete) Hw Score: 70.83%, 4.25 of 6 pt 5.2.14-T Question Help A manufactunng company reglarly ucts quality control chocks at spected periods on the products it manufactures Historically he taure rate for LED light bulbs that the company manufactures is 4% sppose a random sample of 10 LEDigt bulbs is selected Completo parts (a) throug (d) below. The probabety that none of the LED light bulbs are detective s 8648 (Type an rteger or a domal. Round to four domal places as needed ) b. What is the probability that exactly one of the LED light bulbs is defective? The probability that exactly one of the LED light bulbs is defective is 2770 (Type an integer or a decimal. Round to four decimal places as needed ) c. What is the probability that two or fewer of the LED light bubs are defective? The probability that two or fewer of the LED ight bulbs are defective is 9938 Type an integer or a deoimal. Round to four decimal places as needed.) d. What is the probability that three or more of the LED light bulbs are defective? The probabity that three or more of the LED ight bulbs are defective is (Type an integer or a decimal. Round to four decimal places as needed.) Enter your answer in the answer box and then click Check Answer Check Answer to search Home Endb 5 6 7 8 9 0 H J K LExplanation / Answer
Probability of one led being defective is 0.04 ; so probability of one led being non defective is 1-0.04 = 0.96
a) Probability that none of the 10 leds is defective = (0.96)^10 = 0.6648
b) Probability that exactly 1 of the 10 leds is defective is = 10C1 * 0.04 * (0.96)^9 = 0.2770
c) Probability that 2 or fewer leds are defective = probability that none are defective + probability that exactly 1 led is defective + probability that exactly 2 leds are defective = 0.6648 + 0.2770 + 10C2 * (0.04)^2 * (0.96)^8 = 0.9938
d) Probability that 3 or more leds are defective = 1 - probability that 2 or fewer leds are defective = 1 - 0.9938 = 0.0062
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