<p>Luke and Andre are members of a club, which has 8 members. If the club randomly selects 5 members to be on a committee, what is the probability that Andre and Luke are members of the committee?</p>
<p>The probability that Andre is on the committee is 5/8. The probability that Luke is on the committee is 5/8. You know that P(AUB)=P(A)P(B) so the probability they will both be on the committee is 25/64.</p>
<p>You need to look at this in a little more detail. The probability that Andre is on the committee is 5/8. But now there are only 4 remaining slots on the committee, and 7 remaining members of the club, including Luke. (If Andre was not on the committee, which occurs with a probability of 3/8, then they can’t both be on the committee.) So with 4 slots and 7 remaining members, the odds that Luke is on the committee, if you know that Andre is on the committee, are 4/7, and the overall odds are (5 times 4)/(8 times 7). </p>
<p>The reason that RandomHSer’s method does not work is that the events are not independent of each other. Once one committee member has been picked, the number of available slots for the next committee member is reduced.</p>
<p>Another way to think about this: How many total committee are possible? That is given by (8 choose 5), which is 8!/(5! times 3!).
Now, how many of those committees have both Luke and Andre as members? Two slots are tied up by Luke and Andre. Therefore, there are 3 remaining slots on the committee, and there are 6 other members who might fill those slots. The number of committees with both Luke and Andre on them is therefore (6 choose 3), which is 6!/(3! times 3!).
The probability that both Luke and Andre are on the committee is the ratio of the number of committees with both of them as members, to the total number of possible committees. This ought to work out to the same answer as in #3.</p>
<p>Thank you for correcting me. Of course you are right, if one were to carry my solution to its logical extreme, then the probability for a one person committee and a club of two people would be 1/4 when of course it should be zero.</p>
<p>i think it’s 5/14. There is a 5/8 chance that the first person is in the committe. Then there is a 4/7 chance that the second person is in the committee. Multiply them together and you get 5/14.</p>
<p>Right, which would be the same as in #3 and presumably the same as in #4 although I did not compute that explicitly. The SAT probably requires you to reduce the answer to lowest terms, although if a person is wondering about a question of this type, he/she can probably do that!</p>
<p>When you are working on probability problems on the SAT, it is important to consider whether the random events are independent or not. A clearer example occurs when you have an urn/vase/box/bag with different colored marbles, and you are given the number of each. Then you want to determine the probability of certain outcomes when you draw 2 marbles, one at a time. It matters whether you return the first marble that you drew to the group in the urn/vase/box/bag, or whether you keep it out. If you return it, then the draws are independent, and RandomHSers product rule applies. If you keep it out, the first draw changes the number of marbles in the container and the number of the same color as the first. In a problem like this, if you are looking for the probability to draw one red marble and one blue marble, you also need to keep track of the possibility that you draw the red one first or the blue one first.</p>
<p>Thank you all for you efforts :)</p>
<p>I like what @QuantMech proposed. </p>
<p>The number of ways of selecting 5 members from a group of 8 with no restrictions is 8C5 = 8!/(5!)(3!) = 56. </p>
<p>The number of ways of selecting 5 members from a group of 8 including Andre and Luke would mean we already have two people chosen and we just need to select 3 people from the remaining 6 people in the group which is equal to 6C3 or 6!/(3! 3!) = 20 </p>
<p>Answer 20/56 or 5/14. </p>
<p>I do want to point out that this question is beyond the scope of the SAT(not SAT math subject test where this fair game), at least based on all the counting problems I have seen so far on the SAT official tests. Please correct me if you disagree.</p>
<p>Here is a variation:
Luke and Andre are members of a club, which has 8 members. If the club randomly selects 5 members to be on a committee, what is the probability that neither Andre nor Luke are members of the committee? [The answer is not 1 - 5/14]</p>
<p>This might help cement the difference between these three conditions: Andre and Luke, neither Andre nor Luke, Andre or Luke.</p>