Could someone please let me know if I did this problem right?
A bag contains seven blue marbles and five red marbles. One marble is drawn at random without replacement, and then a second marble is drawn. What is the probability that the second marble is blue if the first marble was blue?
I say 6/11, is this right?
Thanks.
If the first drawn marble was blue, there are 6 blues and 5 reds left. Yes, the answer is 6/11.
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To determine the probability that the second marble is blue if the first marble was blue, we can use a conditional probability approach.
First, let's consider the total number of marbles in the bag before any drawing occurs. Since there are 7 blue marbles and 5 red marbles, the total number of marbles is 7 + 5 = 12.
Now, when drawing the first marble, we have 7 blue marbles out of the total 12 marbles. So, the probability of drawing a blue marble as the first marble is 7/12.
After drawing the first blue marble, there are now only 11 marbles left in the bag. Out of these 11 marbles, there are 6 blue marbles remaining. Therefore, the probability of drawing a blue marble as the second marble, given that the first marble was blue, is 6/11.
So, your answer of 6/11 is correct. Well done!