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A bag contains 5 blue marbles, 6 red marbles, and 9 green marbles. Two marbles are drawn at random, one at a time and without replacement. What is the probabili



Please complete the question.  I think you exceeded the character allowance for the question heading.


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3 Answers

Without replacement:

If you want 1 blue marble and 1 red marble, then the probability of 1 blue, followed by 1 red is (5/20)(6/19).

The probability of 1 red, followed by 1 blue is (6/20)(5/19).

So, the probabilty without respect to order is the sum of the 2 scenarios above:  (5/20)*(6/19) + (6/20)*(5/19).

I hope this helps.

Ken L.



Hey Trey -- we can begin even without the entire problem stated (5B,6R,9G)

1st marble: chance of blue= 5/20, 25%; chance of red= 6/20, 30%; green 45%

2nd marble pick: matching blue= 4/19(5/20)= 1/19 or a bit more than 5% ...

                         matching red = 5/19(6/20)= 3/38 or a bit more than 1/13 or 8%

                         matching grn = 8/19(9/20)= 18/95 or ~19% ... no match ~67%

You may adapt the above to address your actual problem ... Best wishes, sir :)

You have 20 marbles in total. The probability to draw a blue marble is 5/20 = 1/4 = 0.25. The probability to draw a red marble is 6/20 = 0.3. The  probability to draw a green marble is 9/20 = 0.45. If two marbles are drawn at random without replacement and one at a time, then the probability that one is red and another is blue is 0.25x0.3 = 0.075. The probability for blue and green is 0.25x0.45 = 0.1125. The probability for red and green is 0.3x0.45=0.135. I think, one of these probabilities is applicable to the situation you  mean in your unfinished statement.