<p>Five marbles are in a bag: are black and are blue. Two marbles are selected without replacement. Given that the second one selected was black, what is the probability that the first was blue?</p>
<p>My answer to that question was .3, but sparknotes said it was .75, but I don't see how they got that answer because their explanation doesn't really make sense....</p>
<p>How many are black and how many are blue?</p>
<p>oh sry, the numbers were loaded up as images...</p>
<p>Five marbles are in a bag: 2 are black and 3 are blue. Two marbles are selected without replacement. Given that the second one selected was black, what is the probability that the first was blue?</p>
<p>mm order doesn't make a difference in this case. You might as well have selected black first, the probability for selecting a blue one will then be 3/4 = 0.75</p>
<p>oh wait, i get it now</p>
<p>man, why was i so stupid??!?!</p>
<p>Can you see the difference with this problem:</p>
<p>Q: A bag contains 3 white marbles and 2 black marbles. Two marbles are drawn from the bag with the first marble not being returned to the bag for the second drawing. What is the probability of getting 1 white marble and 1 black marble?</p>
<p>A: There are 5 marbles total
On the first draw, there is 3/5 chance of getting a white.
On the second draw there are 4 marbles left, and there is a 1/2 chance of getting a black since there are only 2 whites and 2 blacks left.</p>
<p>So 1/2 * 3/5 = 3/10 chance</p>
<p>PS That's what Khalid did the first time.</p>
<p>Is the answer to the initial question 3/5?</p>
<p>whoa whoa whoa, i dont get it
wasnt the question "what is the probability that the first was blue?"
so wouldnt the answer be 3/5, because its asking about the very first marble?</p>
<p>For the original question, it may help to simply write all the options. You NEED to consider the first part of the question, "Given that the second one selected was black, what is the probability that the first was blue?"</p>
<p>Since they TELL you that the second marble is black, here are the options.</p>
<p>U for Blue - A for Black</p>
<p>Picked - Still In the bag
U A ----- A U U
U A ----- U U A
U A ----- U U A
A A ----- U U U </p>
<p>Those are the only options possible. Out of 4 possible outcomes, there could be only 3 successful or 3/4 or .75. </p>
<p>The shortest way to look at it, since we KNOW that the second marble was black, we could reverse the first and second pick and ask ourselves how many are left after picking 1 ... that is 5 minus 1 or 4. But how many blues are left out of the four .... that is 3. The probability to pick a blue marble is thus 3 out of 4. </p>
<p>The reason that the second problem is different is because we have to account for the different probability of drawing a white AND a black marble (since the problem does not tell us that one is established).</p>
<p>Isn't there another option?</p>
<p>U A ----- A U U
U A ----- U U A
U A ----- U U A
A1 A2 ----- U U U
A2 A1-----U U U</p>
<p>Because the question does not specify the specific black marble chosen in the second pick, the possibilities for the first pick can include two different black marble combinations.Is this correct?</p>
<p>yaja,
In your example above, each U A--- line really represents both U A1--- and U A2 ---.<br>
Another way to look at it is to take each possible individual combination of the first two picks (given that a black marble is picked second):
U1A1
U1A2
U2A1
U2A2
U3A1
U3A2
A1A2
A2A1
8 possible combos: 2 black/black, 6 blue/black</p>
<p>Thanks Justadog. Appreciate the help and understand it clearly now:)</p>
<p>Yeah, thanks for the help everyone. That was probably the toughest probability problem that I've seen on an SAT/Practice SAT.</p>