Bunuel wrote:
A bag contains 3 white, 4 black, and 2 red marbles. Two marbles are drawn from the bag. If replacement is not allowed, what is the probability that the second marble drawn will be red?
A. \(\frac{1}{36}\)
B. \(\frac{1}{12}\)
C. \(\frac{7}{36}\)
D. \(\frac{2}{9}\)
E. \(\frac{7}{9}\)
This question reminds me of my childhood, when my friends and I would sometimes "draw straws" to randomly select one person to do something (often either work, like getting wood for the fire, or dumb, like eating something that shouldn't be eaten).
So, someone would hold up n pieces of grass (for n people), and one of those pieces was very short. The person who selected the shortest piece was the one who had to perform the task.
There was always one guy who wanted to choose his piece last. His reasoning was that his chances of drawing the shortest piece would be minimized, since every person before him had a chance of drawing the short piece before it got to his turn.
The truth of the matter is that each of the n people had a 1/n chance of selecting the shortest piece,
regardless of the order in which they selected.
The same applies to the original question here.
2 of the 9 marbles are red, so P(red ball selected SECOND) = P(red ball selected FIRST) = 2/9
Cheers,
Brent
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