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A lottery is played by selecting X distinct single digit numbers from
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05 Feb 2013, 07:40
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A lottery is played by selecting X distinct single digit numbers from 0 to 9 at once such that order does not matter. What is the probability that a player will win playing the lottery? (1) Players must match at least two numbers with machine to win. (2) X = 4
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Re: A lottery is played by selecting X distinct single digit numbers from
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30 Jun 2017, 12:07
TheMastermind wrote: Excellent explanation. I was wondering how can we calculate the probability for winning the lottery with the given information. We know it's sufficient but what's the actual solution if this question was a PS problem.
Thank you. Dear TheMastermind, I'm happy to respond. Just so we are clear, here would be the PS version of the problem: A lottery is played by selecting 4 distinct single digit numbers from 0 to 9 at once such that order does not matter. The lottery machine also picks four digits in the same way. Players must match at least two numbers with the machine to win. What is the probability that a player will win playing the lottery? Think about it this way. I pick four of the digits at random. That's fixed. Now, the lottery machine is going to pick four digits at random. Here, I will use "at least" logic in the calculation. That is, for ease and convenient, we will calculate the probabilities of all the cases not included in the solution. Case I: the machine's four choices don't overlap at all with mine. There are four digits, my chosen digits, that the machine can't pick. The machine has to pick from the remaining six digits: first choice P = 6/10 = 3/5 second choice P = 5/9 third choice P = 4/8 = 1/2 fourth choice = 3/7 \(P_1 = \dfrac{3}{5} \times \dfrac{5}{9} \times \dfrac{1}{2} \times \dfrac{3}{7}\) \(P_1 = \dfrac{1}{1} \times \dfrac{1}{1} \times \dfrac{1}{2} \times \dfrac{1}{7} = \dfrac{1}{14}\) That's the Case I probability. Case II: the machine's four choices overlap with exactly one digit of mine. First, there's the one digit that matches one of mine: this could be made on any of the four choices. The other three choices follow the pattern above. The choices of the matching digit in the terms below is the first in the first line, the second in the second line, etc. \(P_2 = \dfrac{4}{10} \times \dfrac{6}{9} \times \dfrac{5}{8} \times \dfrac{4}{7}\) \(+ \dfrac{6}{10} \times \dfrac{4}{9} \times \dfrac{5}{8} \times \dfrac{4}{7}\) \(+ \dfrac{6}{10} \times \dfrac{5}{9} \times \dfrac{4}{8} \times \dfrac{4}{7}\) \(+ \dfrac{6}{10} \times \dfrac{5}{9} \times \dfrac{4}{8} \times \dfrac{4}{7}\) \(P_2 = \dfrac{1}{1} \times \dfrac{2}{3} \times \dfrac{1}{1} \times \dfrac{1}{7}\) \(+ \dfrac{2}{1} \times \dfrac{1}{3} \times \dfrac{1}{1} \times \dfrac{1}{7}\) \(+ \dfrac{1}{1} \times \dfrac{1}{3} \times \dfrac{1}{1} \times \dfrac{2}{7}\) \(+ \dfrac{1}{1} \times \dfrac{1}{3} \times \dfrac{1}{1} \times \dfrac{2}{7}\) \(P_2 = \dfrac{4*2}{3*7} = \dfrac{8}{21}\) That's the Case II probability. Add those two: \(P_{1,2} = \dfrac{1}{14} + \dfrac{8}{21} = \dfrac{3}{42} + \dfrac{16}{42} = \dfrac{19}{42}\) That's the complete probability of condition of the question not being satisfied, so we subtract that from 1 to answer the question. \(P = 1  \dfrac{19}{42} = \dfrac{23}{42}\) That's the answer to the question. This would be an extremely challenging PS. So far as I can tell, there's not a significantly short solution than the one I have shown. Does all this make sense? Mike
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Re: A lottery is played by selecting X distinct single digit numbers from
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05 Feb 2013, 10:12
danzig wrote: A lottery is played by selecting X distinct single digit numbers from 0 to 9 at once such that order does not matter. What is the probability that a player will win playing the lottery? (1) Players must match at least two numbers to win. (2) X = 4 I'm happy to help. This is a somewhat offbeat question, but then again, that's just what the GMAT will throw at you. So, from the prompt, we know we are picking X different single digit numbers: X must be greater than two and less than 9 (or 10, if we are counting zero as a "single digit number"  let's ignore that complication). Order doesn't matter. We know nothing about what constitutes winning. Statement #1: Players must match at least two numbers to win. Now, at least we know what constitutes winning. The trouble is  we don't know the how many digits are picked. If X = 9  the lottery picks all the digits from 19, then I also pick all the digits from 19  then I have 100% chance of matching at least two digits and winning. That wouldn't be much of a lottery. If X = 3  the lottery picks three, and then I pick three  well, that's harder. Clearly the probability of winning depends on the value of X, and we don't know that in Statement #1. This statement, alone and by itself, is insufficient. Statement #2: X = 4Now, we know how many digits are picked  lottery picks 4, then I pick 4  but now I have no idea what constitutes "winning". (This is an example of a DS question in which it's crucially important to forget all about Statement #1 when we are analyzing Statement #2 on its own.) In Statement #2, we know how many digits are picked, but we have absolutely no idea what constitutes winning. This statement, alone and by itself, is insufficient. Combined statements: Now, we know  the lottery picks 4 digits, then I pick 4 digits, and if at least two of my digits match two of the lottery's cards, I win. This is now a well defined math problem, and if we wanted, we could calculate the numerical value of the probability. Of course, since this is DS, it would be a big mistake to waste time with that calculation. We have enough information now. Combined, the statements are sufficient. Answer = CDoes all this make sense? Mike
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Re: A lottery is played by selecting X distinct single digit numbers from
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05 Feb 2013, 11:49
Hi, I think that Mike's explanation is great. I'm just wondering where this question came from? The wording seems strange to the point that the question may not be solvable. Here is the potentially problematic wording: distinct single digit numbers from 0 to 9 What does that mean? Does that mean single digit INTEGERS between 0 and 9 or just single digit numbers. My assumption is that single digit numbers between 0 and 9 are not just "the digits" but could include the single digit decimals .1,.2,.3.... Anyone? I guess that if you assume that the question includes the extra decimals then Mike's solution still works although I'm curious about whether this is an official question. HG.
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Re: A lottery is played by selecting X distinct single digit numbers from
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07 Feb 2013, 08:22
HerrGrau wrote: Hi,
I think that Mike's explanation is great. I'm just wondering where this question came from? The wording seems strange to the point that the question may not be solvable. Here is the potentially problematic wording: distinct single digit numbers from 0 to 9
What does that mean? Does that mean single digit INTEGERS between 0 and 9 or just single digit numbers. My assumption is that single digit numbers between 0 and 9 are not just "the digits" but could include the single digit decimals .1,.2,.3.... Anyone?
I guess that if you assume that the question includes the extra decimals then Mike's solution still works although I'm curious about whether this is an official question.
HG. It's tagged as a grockit problem, so I would guess that this is not an official question. It is also tagged as a 600700 lvl question, which seems high to me for this question...I don't have any experience with Grockit, but MGMAT questions that are in the 600700 range tend to be more difficult for me to solve.



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Re: A lottery is played by selecting X distinct single digit numbers from
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07 Feb 2013, 10:09
HerrGrau wrote: Hi, The wording seems strange to the point that the question may not be solvable. Here is the potentially problematic wording: distinct single digit numbers from 0 to 9 What does that mean? Does that mean single digit INTEGERS between 0 and 9 or just single digit numbers. My assumption is that single digit numbers between 0 and 9 are not just "the digits" but could include the single digit decimals .1,.2,.3.... Anyone? Dear HerrGrauThank you for your kind words. In my experience, I have never heard " single digit numbers" apply to anything other than nonnegative integers. I have never seen any book or hear anyone ever refer to, say, 5 or 0.008 or 3 x 10^8 as a "single digit number. Yes, technically, each of these is written with a single nonzero digit, but I have never heard the term "single digit number" used for them. (We could say that each of these has "one significant figure", but that carries us far afield into measurement theory, well beyond GMAT territory.) In my mind, the only ambiguity in this question is whether zero is included  I guess I was assuming, whatever their understanding of the term, this understanding would be fixed and wouldn't change as we moved through the statements of the DS  therefore, with both statements, and with whatever convention they are following, the question can be definitely answered. I agree with you  this question is not written with the tight precision so characteristic of official GMAT questions. The very fact that there's any ambiguity at all makes this a woefully substandard question. Mike
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Re: A lottery is played by selecting X distinct single digit numbers from
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07 Feb 2013, 14:47
Hi Mike, In general I agree with everything that you're saying. Just to clarify, I think that .008 is a three digit number with 2 unique digits and so would be excluded. The numbers in question are .1, .2, .3, .4, .5, .6, .7, .8, and .9. They are numbers that have only one digit and hence are "single digit numbers". I have never seen this distinction be an actual issue on a GMAT question but am curious as to what the general consensus is. HG.
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Re: A lottery is played by selecting X distinct single digit numbers from
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16 Oct 2014, 11:04
just for the sake of understanding, how would we solve this question? 1 probability of matching none  prob of matching 1? none= (5/9)*4/9*3/9*2/9
probability of matching one: 1*5/9*4/9*3/9
is this the way?
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Re: A lottery is played by selecting X distinct single digit numbers from
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05 Dec 2015, 11:47
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. A lottery is played by selecting X distinct single digit numbers from 0 to 9 at once such that order does not matter. What is the probability that a player will win playing the lottery? 1) Players must match at least two numbers to win. (2) X = 4 There are 2 variables (x and how many to be win) and 2 equations are given, so there is high chance (C) will be the answer. Looking at the conditions together, the probability to win is 1(10C4/10^4), which is unique and makes the condition sufficient, and the answer becomes (C). For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: A lottery is played by selecting X distinct single digit numbers from
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30 Jun 2017, 04:06
mikemcgarry wrote: danzig wrote: A lottery is played by selecting X distinct single digit numbers from 0 to 9 at once such that order does not matter. What is the probability that a player will win playing the lottery? (1) Players must match at least two numbers to win. (2) X = 4 I'm happy to help. This is a somewhat offbeat question, but then again, that's just what the GMAT will throw at you. So, from the prompt, we know we are picking X different single digit numbers: X must be greater than two and less than 9 (or 10, if we are counting zero as a "single digit number"  let's ignore that complication). Order doesn't matter. We know nothing about what constitutes winning. Statement #1: Players must match at least two numbers to win. Now, at least we know what constitutes winning. The trouble is  we don't know the how many digits are picked. If X = 9  the lottery picks all the digits from 19, then I also pick all the digits from 19  then I have 100% chance of matching at least two digits and winning. That wouldn't be much of a lottery. If X = 3  the lottery picks three, and then I pick three  well, that's harder. Clearly the probability of winning depends on the value of X, and we don't know that in Statement #1. This statement, alone and by itself, is insufficient. Statement #2: X = 4Now, we know how many digits are picked  lottery picks 4, then I pick 4  but now I have no idea what constitutes "winning". (This is an example of a DS question in which it's crucially important to forget all about Statement #1 when we are analyzing Statement #2 on its own.) In Statement #2, we know how many digits are picked, but we have absolutely no idea what constitutes winning. This statement, alone and by itself, is insufficient. Combined statements: Now, we know  the lottery picks 4 digits, then I pick 4 digits, and if at least two of my digits match two of the lottery's cards, I win. This is now a well defined math problem, and if we wanted, we could calculate the numerical value of the probability. Of course, since this is DS, it would be a big mistake to waste time with that calculation. We have enough information now. Combined, the statements are sufficient. Answer = CDoes all this make sense? Mike Excellent explanation. I was wondering how can we calculate the probability for winning the lottery with the given information. We know it's sufficient but what's the actual solution if this question was a PS problem. Thank you.



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Re: A lottery is played by selecting X distinct single digit numbers from
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14 Nov 2018, 17:47
mikemcgarry wrote: TheMastermind wrote: Excellent explanation. I was wondering how can we calculate the probability for winning the lottery with the given information. We know it's sufficient but what's the actual solution if this question was a PS problem.
Thank you. Dear TheMastermind, I'm happy to respond. Just so we are clear, here would be the PS version of the problem: A lottery is played by selecting 4 distinct single digit numbers from 0 to 9 at once such that order does not matter. The lottery machine also picks four digits in the same way. Players must match at least two numbers with the machine to win. What is the probability that a player will win playing the lottery? Think about it this way. I pick four of the digits at random. That's fixed. Now, the lottery machine is going to pick four digits at random. Here, I will use "at least" logic in the calculation. That is, for ease and convenient, we will calculate the probabilities of all the cases not included in the solution. Case I: the machine's four choices don't overlap at all with mine. There are four digits, my chosen digits, that the machine can't pick. The machine has to pick from the remaining six digits: first choice P = 6/10 = 3/5 second choice P = 5/9 third choice P = 4/8 = 1/2 fourth choice = 3/7 \(P_1 = \dfrac{3}{5} \times \dfrac{5}{9} \times \dfrac{1}{2} \times \dfrac{3}{7}\) \(P_1 = \dfrac{1}{1} \times \dfrac{1}{1} \times \dfrac{1}{2} \times \dfrac{1}{7} = \dfrac{1}{14}\) That's the Case I probability. Case II: the machine's four choices overlap with exactly one digit of mine. First, there's the one digit that matches one of mine: this could be made on any of the four choices. The other three choices follow the pattern above. The choices of the matching digit in the terms below is the first in the first line, the second in the second line, etc. \(P_2 = \dfrac{4}{10} \times \dfrac{6}{9} \times \dfrac{5}{8} \times \dfrac{4}{7}\) \(+ \dfrac{6}{10} \times \dfrac{4}{9} \times \dfrac{5}{8} \times \dfrac{4}{7}\) \(+ \dfrac{6}{10} \times \dfrac{5}{9} \times \dfrac{4}{8} \times \dfrac{4}{7}\) \(+ \dfrac{6}{10} \times \dfrac{5}{9} \times \dfrac{4}{8} \times \dfrac{4}{7}\) \(P_2 = \dfrac{1}{1} \times \dfrac{2}{3} \times \dfrac{1}{1} \times \dfrac{1}{7}\) \(+ \dfrac{2}{1} \times \dfrac{1}{3} \times \dfrac{1}{1} \times \dfrac{1}{7}\) \(+ \dfrac{1}{1} \times \dfrac{1}{3} \times \dfrac{1}{1} \times \dfrac{2}{7}\) \(+ \dfrac{1}{1} \times \dfrac{1}{3} \times \dfrac{1}{1} \times \dfrac{2}{7}\) \(P_2 = \dfrac{4*2}{3*7} = \dfrac{8}{21}\) That's the Case II probability. Add those two: \(P_{1,2} = \dfrac{1}{14} + \dfrac{8}{21} = \dfrac{3}{42} + \dfrac{16}{42} = \dfrac{19}{42}\) That's the complete probability of condition of the question not being satisfied, so we subtract that from 1 to answer the question. \(P = 1  \dfrac{19}{42} = \dfrac{23}{42}\) That's the answer to the question. This would be an extremely challenging PS. So far as I can tell, there's not a significantly short solution than the one I have shown. Does all this make sense? Mike Hi Mike, I know this is DS, so we do not need to solve, but I was able to come up with the same solution, using the following method  is there anything done incorrectly? 1. Total possible machine selections: 10C4 = 210 2. Total possible human selections where no numbers match (choose any 4 from the remaining 6 numbers the machine did not select): 6C4 = 15 3. Total possible human selections where 1 number matches (choose one of the 4 numbers selected by machine x 3 of the unselected numbers): 4 x 6C3 = 4 x 20 = 80 Probability that human does not win: 1 (combined losing selections/total possible machine selections) = 1  ((15+80)/210) = 1  19/42 = 23/42




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