Asked by Matt
Sorry here is completed question
A bag contains 3 red marbles, 3 blue marbles and 3 yellow marbles. Which of the following methods of selecting 2 marbles will have the greater probability of choosing 2 red marbles?
1- Randomly choose a marble don't replace it then randomly choose another marble
2 - Randomly choose a marble replace it then randomly choose another marble
A bag contains 3 red marbles, 3 blue marbles and 3 yellow marbles. Which of the following methods of selecting 2 marbles will have the greater probability of choosing 2 red marbles?
1- Randomly choose a marble don't replace it then randomly choose another marble
2 - Randomly choose a marble replace it then randomly choose another marble
Answers
Answered by
Steve
You could have added a posting to the original question. Anyway,
There are 3 red marbles out of 9 total, so on the first draw,
P(red) = 3/9 = 1/3
Without replacement, there are now 2 of 8 reds, so P(red) = 2/8 = 1/4
P(red,red) = 1/3 * 1/4 = 1/12
With replacement, both draws are like the first, so
P(red,red) = 1/3 * 1/3 = 1/9
There are 3 red marbles out of 9 total, so on the first draw,
P(red) = 3/9 = 1/3
Without replacement, there are now 2 of 8 reds, so P(red) = 2/8 = 1/4
P(red,red) = 1/3 * 1/4 = 1/12
With replacement, both draws are like the first, so
P(red,red) = 1/3 * 1/3 = 1/9
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