Hello,
The correct answer is C. However, I don’t understand why probability of the man pulls out a read marble fourth is equal to the probability of the man pulls out a green marble first. I show how I calculated this question in the picture below. I wonder what’s wrong with my steps. Thank you.
Your Quantity A is the probability that the man pulls out a red marble four times in a row, not the fourth time.
1 Like
I multiply probabilities of drawing the first red marble, the second red marble, and third red marble and the forth, and the result doesn’t equal to 1/ 2 which is the probability of drawing the first marble is green. I wonder what’s wrong with my steps. Thank you.
That’s not what the question is asking though. It’s asking “on the fourth pull, what’s the probability that the man gets a red ball?”. This is different because you can have the first three pulls be green and then the fourth - in other words we don’t know what the result of the first three pulls would be.
So what should I do ? Thanks.
Try with a smaller example. Suppose you have five red marbles and 6 green marbles. What if Quantity A was “second” and Quantity B “first” instead?
If Quantity A is second, then the probability of drawing second red marble should be 5 / 16 multiply by 4 / 15. If Quantity B is first, then the probability of drawing first green is 6 / 11.
That’s the problem. For Quantity A: we have two cases:
- we get a red marble first and a red marble second
- we get a green marble first and a red marble second
The probability of the first case occurring is
\frac{5}{11} \times \frac{4}{10}
The probability of the second case occurring is
\frac{6}{11} \times \frac{5}{10}
The total probability is
\frac{5}{11} \times \frac{4}{10} + \frac{6}{11} \times \frac{5}{10}
=\frac{50}{110} = \frac{5}{11}
which is less than Quantity B (which you correctly found out to be \frac{6}{11}). Does that help?
Hello. Thank you so much for your help. I understand this question. However, I think that the original question is way more complicated. If I want to calculate the probability of drawing fourth red marble, I have to calculate the probability of the three draws before the fourth one. There are multiple possibilities of the first three draws. For example, the first three draws could be all red marbles or all green marbles. In this case, I wonder how Vince could solve this question without any calculation. Thanks.
A user has formally explained why all of this doesn’t really matter.