H2 Math: Conditional Probability Question

• Lord dejavu

  19 Jul 10, 19:26

  There are 4 red and 6 green balls in a bag. Four are drawn at radom w/o replacement. What is the probability that at least two are geen, given that at least one of each color is drawn? I tried to use P(.....) = P(X n Y) / P(Y) , but i dont know how to find X intersect Y, which in this case is, more than two green and one red ball are chosen. Is it possible to solve this question without using Permutation and Combination commands? ie n C x etc. thanks.
• Mad Hat

  19 Jul 10, 21:31

  Yes, can do directly:

  P(At least 2G / At least 1R,1G)

  = P(At least 2G and At least 1R,1G) ÷ P(At least 1R,1G)

  = P(At least 2G,1R) ÷ P(At least 1R,1G)

  = P( 3G,1R  or  2G,2R ) ÷ P(At least 1R,1G)

  = { P(3G,1R) + P(2G,2R) } ÷ P(At least 1R,1G)

  So you just need to find P(3G,1R), P(2G,2R) and P(At least 1R,1G).