JAMB - Mathematics (1985 - No. 37)
A bag contains 4 white balls and 6 red balls. Two balls are taken from the bag without replacement. What is the probability that they are both red?
\(\frac{2}{3}\)
\(\frac{2}{15}\)
\(\frac{1}{2}\)
\(\frac{1}{3}\)
Explanation
P(R1) = \(\frac{6}{10}\)
= \(\frac{2}{3}\)
P(R1 n R11) = P(both red)
\(\frac{3}{5}\) x \(\frac{5}{9}\)
= \(\frac{1}{3}\)
= \(\frac{2}{3}\)
P(R1 n R11) = P(both red)
\(\frac{3}{5}\) x \(\frac{5}{9}\)
= \(\frac{1}{3}\)
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