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})