Bowl A contains 6 red chips and 4 blue chips Four of these 1

Bowl A contains 6 red chips and 4 blue chips. Four of these 10 chips are selected at
random and without replacement and put into bowl B, which contains 2 blue chips
originally. One chip is then drawn at random from bowl B. If a red chip is selected,
what is the probability that 2 red chips and 2 blue chips were transferred from bowl A
to bowl B?

Solution

n(s)= number of way of drawing 4 chips out of 10 chips =10C4=(10*9*8*7)/(4*3*2)=210. let e be the probability of drawing 2 red chips and 2 blue chips. therefore n(E)= number of way of drawing 2 red chips out of 6 red chips and 2 blue chips out of 4 blue chips. =6C2 * 4C2=15*6=90. So, the required probability P(E)=n(E)/n(s)=90/210=3/7 (Ans) htpp//:tutorteddy.com