A bag contains 8 red, 5 white and 7 blue balls. A person picks three balls at random. Find the difference between the probability that out of three, 2 balls are of red, and out of three, at least two balls are blue.
Solution
Total number of balls in the bag = 8 + 5 + 7 = 20
Probability that out of three, 2 balls are red
$= \dfrac{(^8C_2 \times ^{12}C_1)}{20^C_3} = \dfrac{(28 \times 12)}{1140} = \dfrac{336}{1140}$
Probability that out of three, at least two balls are blue = $\dfrac{\left[(^7C_2 \times ^{13}C_1) + ^7C_3 \right]}{^{20}C_3}$
$= \dfrac{\left[(21 \times 13) + 35\right]}{1140} = \dfrac{308}{1140}$
Required difference $= \dfrac{336}{1140} – \dfrac{308}{1140}$
$= \dfrac{28}{1140} = \dfrac{7}{285}$