# SOLUTION: Suppose an opaque jar contains 5 red marbles and 10 green marbles. I experiment picking 2 marbles from the jar without replacing the first one. 1. What is the probability of get

Algebra ->  Algebra  -> Probability-and-statistics -> SOLUTION: Suppose an opaque jar contains 5 red marbles and 10 green marbles. I experiment picking 2 marbles from the jar without replacing the first one. 1. What is the probability of get      Log On

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 Question 593109: Suppose an opaque jar contains 5 red marbles and 10 green marbles. I experiment picking 2 marbles from the jar without replacing the first one. 1. What is the probability of getting a green marble first and a red marble second? 2. What is the probability of getting a green marble and a red marble? How is this different from question #1? For the 1st question I did this: (5x10)/(15x14)= 50/210 = 5/21 (am pretty sure that is right) For the 2nd question I am stumped...I thought it would be the same answer but something keeps nagging at me that I am missing something. I thought about this: (5)squared +(10)squared/(15x14) = 125/210.595 or 59.5% I really don't understand how the questions are that different..or at least I know they are asking 2 different probable outcomes but don't know how to explain the difference. Can you please help me? Thanks! Glennys Answer by stanbon(57347)   (Show Source): You can put this solution on YOUR website!Suppose an opaque jar contains 5 red marbles and 10 green marbles. I experiment picking 2 marbles from the jar without replacing the first one. 1. What is the probability of getting a green marble first and a red marble second? Ans: (10/15)(5/14) = 50/210 = 5/21 ---- 2. What is the probability of getting a green marble and a red marble? How is this different from question #1? # of ways to succeed: 5*10 = 50 # of random pairs: 15C2 = 105 --- P(red AND green) = 50/105 = 10/21 ============= Cheers, Stan H. =============