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Of the 20 lightbulbs in a box, 2 are defective. An inspector will select 2 light

游客2024-01-12  66
问题 Of the 20 lightbulbs in a box, 2 are defective. An inspector will select 2 lightbulbs simultaneously and at random from the box. What is the probability that neither of the lightbulbs selected will be defective?
Give your answer as a fraction.
选项
答案 153/190
解析 The desired probability corresponds to the fraction
the number of ways that 2 lightbulbs, both of which are not defective, can be chosen
            the number of ways that 2 lightbulbs can be chosen
    In order to calculate the desired probability, you need to calculate the values of the numerator and the denominator of this fraction.
    In the box there are 20 lightbulbs, 18 of which are not defective. The numerator of the fraction is the number of ways that 2 lightbulbs can be chosen from the 18 that are not defective, also known as the number of combinations of 18 objects taken 2 at a time.
    If you remember the combinations formula, you know that the number of combinations is(which is denoted symbolically asor). Simplifying, you get

    Similarly, the denominator of the fraction is the number of ways that 2 lightbulbs can be chosen from the 20 in the box, which is . Therefore, the probability that neither of the lightbulbs selected will be defective is 153/190. The correct answer is 153/190(or any equivalent fraction).
    Another approach is to look at the selection of the two lightbulbs separately. The problem states that lightbulbs are selected simultaneously. However, the timing of the selection only ensures that the same lightbulb is not chosen twice. This is equivalent to choosing one lightbulb first and then choosing a second lightbulb without replacing the first. The probability that the first lightbulb selected will not be defective is 18/20. If the first lightbulb selected is not defective, there will be 19 lightbulbs left to choose from, 17 of which are not defective. Thus, the probability that the second lightbulb selected will not be defective is 17/19. The probability that both lightbulbs selected will not be defective is the product of these two probabilities. Thus, the desired probability is . The correct answer is 153/190(or any equivalent fraction).
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