WAEC - Mathematics (2016 - No. 41)
A bag contains 5 red and 4 blue identical balls. Id two balls are selected at random from the bag, one after the other, with replacement, find the probability that the first is red and the second is blue
\(\frac{2}{9}\)
\(\frac{5}{18}\)
\(\frac{20}{81}\)
\(\frac{5}{9}\)
Explication
n(red balls) = 5
n(blue balls) = 4
n(\(\iff\)) = 9
Hence, prob (R1, B2)
= \(\frac{5}{9} \times \frac{4}{9}\)
= \(\frac{20}{81}\)
n(blue balls) = 4
n(\(\iff\)) = 9
Hence, prob (R1, B2)
= \(\frac{5}{9} \times \frac{4}{9}\)
= \(\frac{20}{81}\)