A bag contains 12 white balls and 8 black balls, another contains 10 white balls
A bag contains 12 white balls and 8 black balls, another contains 10 white balls and 15 black balls. If two balls are drawn, without replacement from each bag, find the probability that :
(a) all four balls are black ;
(b) exactly one of the four balls is white.
Explanation
(a) 1st bag contains 12 white (W) balls and 8 black (B) balls; 2nd bag contains 10 white (W) balls and 15 black (B) balls.
P(two black balls from the 1st bag and two black balls from 2nd bag)
= \((\frac{8}{20} \times \frac{7}{19}) \times (\frac{15}{25} \times \frac{14}{24})\)
= \(\frac{14}{95} \times \frac{7}{20}\)
= \(\frac{49}{950}\)
(b) P(picking exactly one white ball) = P(WB from 1st bag and BB from 2nd bag) + P(BW from 1st bag and BB from 2nd bag) + P(BB from 1st bag and WB from 2nd bag) + P(BB from 1st bag and BW from 2nd bag)
= \(\frac{12}{20} \times \frac{8}{19} \times \frac{15}{25} \times \frac{14}{24} + \frac{8}{20} \times \frac{12}{19} \times \frac{15}{25} \times \frac{14}{24} + \frac{8}{20} \times \frac{7}{19} \times \frac{10}{25} \times \frac{15}{24} + \frac{8}{20} \times \frac{7}{19} \times \frac{15}{25} \times \frac{10}{24}\)
= \(\frac{42}{475} + \frac{42}{475} + \frac{7}{190} + \frac{7}{190}\)
= \(\frac{238}{950}\)