Mathematics: Suppose a box contains 3 defective light bulbs and 12 good bulbs. Two bulbs are chosen separately from the box without replacement. What is the probability that one of two bulbs drawn is defective and one is not?

7 bowls Step-by-step explanation:5_ 5/6 divided by 3/4 equals 7.77777...

Suppose a box contains 3 defective light bulbs

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09.02.2022

$Probability = \frac{12}{35}$

Step-by-step explanation:

Given

Represent Defective Bulbs with D and non good bulbs with G

D = 3

G = 12

The probability that one is defective and the other is not can be represented with: (D and G) or (G and D)

Which means that the defective is selected first, then the good one or the good one is selected first, then the defective one.

This is then calculated as:

$Probability = \frac{n(D}{Total} * \frac{n(G)}{Total - 1} + \frac{n(G}{Total} * \frac{n(D)}{Total - 1}$

We used Total - 1 on second selections because it is a probability without replacement

$Probability = \frac{3}{15} * \frac{12}{15- 1} + \frac{12}{15} * \frac{3}{15- 1}$

$Probability = \frac{3}{15} * \frac{12}{14} + \frac{12}{15} * \frac{3}{14}$

$Probability = \frac{1}{5} * \frac{6}{7} + \frac{2}{5} * \frac{3}{7}$

$Probability = \frac{6}{35} + \frac{6}{35}$

$Probability = \frac{12}{35}$

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