Answer:
Pr(of getting 1grape ball, 1 cherry ball and 1 lemon ball) =0.20832
Step-by-step explanation:
Let the grape falvored ball be represented by G,
Cherry flavored balls be C
Lemon flavored balls be L
the possible order of picks for getting 1grape ball, 1 cherry ball and 1lemon ball are; GCL or GLC or CGL or CLG or LGC or LCG
Pr(of getting 1grape ball, 1 cherry ball and 1 lemon ball)
= Pr(GCL) or Pr(GLC) or Pr(CGL) or Pr(CLG) or Pr(LGC) or Pr(LCG)
with replacement we have;
Probability = [tex]\frac{number of required outcomes}{number of possible outcomes}[/tex]
= [tex]\frac{300}{1200}*\frac{400}{1200}*\frac{500}{1200} +\frac{300}{1200}*\frac{500}{1200}*\frac{400}{1200} +\frac{400}{1200}*\frac{300}{1200}*\frac{500}{1200} \\ \\+\frac{400}{1200}*\frac{500}{1200}*\frac{300}{1200} +\frac{500}{1200}*\frac{300}{1200}*\frac{400}{1200} +\frac{500}{1200}*\frac{400}{1200}*\frac{300}{1200}[/tex]
= 0.03472 + 0.03472 +0.03472 + 0.03472 + 0.03472 + 0.03472
=0.20832
