{"id":20309,"date":"2024-04-15T05:51:10","date_gmt":"2024-04-15T05:51:10","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=20309"},"modified":"2024-04-15T05:51:10","modified_gmt":"2024-04-15T05:51:10","slug":"there-are-25-calculators-in-a-box-two-of-them-are-defective-suppose-5-calculators-are-randomly-picked-for-inspection-i-e-each-has-the-same-chance-of-being-selected-what-is-the-probability-that","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/there-are-25-calculators-in-a-box-two-of-them-are-defective-suppose-5-calculators-are-randomly-picked-for-inspection-i-e-each-has-the-same-chance-of-being-selected-what-is-the-probability-that\/","title":{"rendered":"There are 25 calculators in a box. Two of them are defective. Suppose 5 calculators are randomly picked for inspection (i.e., each has the same chance of being selected), what is the probability that only one of the defective calculators will be included in the inspection? A. $$\\frac{1}{2}$$ B. $$\\frac{1}{3}$$ C. $$\\frac{1}{4}$$ D. $$\\frac{1}{5}$$"},"content":{"rendered":"

[amp_mcq option1=”$$\\frac{1}{2}$$” option2=”$$\\frac{1}{3}$$” option3=”$$\\frac{1}{4}$$” option4=”$$\\frac{1}{5}$$” correct=”option3″]<\/p>\n

The correct answer is $\\boxed{\\frac{1}{20}}$.<\/p>\n

There are two ways to choose one defective calculator and four good calculators:<\/p>\n

    \n
  1. Choose one defective calculator from the two defective calculators and then choose four good calculators from the 23 good calculators. This can be done in $\\binom{2}{1}\\binom{23}{4} = \\frac{25 \\times 24 \\times 23 \\times 22}{4 \\times 3 \\times 2 \\times 1} = 2300$ ways.<\/li>\n
  2. Choose four good calculators from the 23 good calculators and then choose one defective calculator from the two defective calculators. This can be done in $\\binom{23}{4}\\binom{2}{1} = \\frac{23 \\times 22 \\times 21 \\times 20}{4 \\times 3 \\times 2 \\times 1} = 1380$ ways.<\/li>\n<\/ol>\n

    Therefore, the probability of choosing one defective calculator and four good calculators is $\\frac{2300 + 1380}{25 \\times 24} = \\frac{3680}{600} = \\boxed{\\frac{1}{20}}$.<\/p>\n

    Option A is incorrect because it is the probability of choosing two defective calculators. Option B is incorrect because it is the probability of choosing three defective calculators. Option C is incorrect because it is the probability of choosing four defective calculators. Option D is incorrect because it is the probability of choosing none of the defective calculators.<\/p>\n","protected":false},"excerpt":{"rendered":"

    [amp_mcq option1=”$$\\frac{1}{2}$$” option2=”$$\\frac{1}{3}$$” option3=”$$\\frac{1}{4}$$” option4=”$$\\frac{1}{5}$$” correct=”option3″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[691],"tags":[],"class_list":["post-20309","post","type-post","status-publish","format-standard","hentry","category-probability-and-statistics","no-featured-image-padding"],"yoast_head":"\nThere are 25 calculators in a box. Two of them are defective. Suppose 5 calculators are randomly picked for inspection (i.e., each has the same chance of being selected), what is the probability that only one of the defective calculators will be included in the inspection? 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