Three red balls, five green balls, and a number of blue balls are put together in a sack. One ball is picked at random from the sack. If the probability of picking a red ball is \(\frac{1}{6}\) find;
(a) The number of blue balls in the sack
(b) the probability of picking a green ball
(a) Let number of blue balls = n
Total number of balls = 3 + 5 + n
= 8 + n
p(red ball) = \(\frac{\text{no. of red balls}}{\text{Total no. of balls}}\)
\(\frac{1}{6} = \frac{3}{8 + n}\)
8 + n = 18
n + 18 - 8 = 10
number of blue balls = 10
(b) P(green ball) = \(\frac{\text{No. of green balls}}{\text{Total no. of balls}}\)
= \(\frac{5}{3 + 5 + 10}\)
= \(\frac{5}{18}\)
= 0.2778
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