Three defective bulbs got mixed up with seven good ones. If two bulbs are selected at random, what is the probability that both are good?
\(\frac{3}{7}\)
\(\frac{21}{50}\)
\(\frac{7}{15}\)
\(\frac{49}{100}\)
Correct answer is C
\(P(\text{both bulbs are good}) = \frac{7}{10} \times \frac{6}{9}\)
= \(\frac{7}{15}\)