WAEC - Mathematics (2012 - No. 20)
A bag contains 4 red and 6 black balls of the same size. If the balls are shuffled briskly and two balls are drawn one after the other without replacement, find the probability of picking balls of different colours
\(\frac{8}{15}\)
\(\frac{13}{25}\)
\(\frac{11}{15}\)
\(\frac{13}{15}\)
Wyjaśnienie
Prob(RB + BR) = Total balls = 4 + 6 = 10
= prob(\(\frac{4}{10} \times \frac{6}{9}\)) + prob(\(\frac{6}{10} \times \frac{4}{9}) = \frac{24}{90} + \frac{24}{90}\)
= \(\frac{48}{90} = \frac{16}{30} = \frac{8}{15}\)
= prob(\(\frac{4}{10} \times \frac{6}{9}\)) + prob(\(\frac{6}{10} \times \frac{4}{9}) = \frac{24}{90} + \frac{24}{90}\)
= \(\frac{48}{90} = \frac{16}{30} = \frac{8}{15}\)