A lottery is played by selecting X distinct single digit numbers from

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.

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?

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 1-9, then I also pick all the digits from 1-9 --- 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 = 4
Now, 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 = C

Does all this make sense?

Mike

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

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,
Can you explain why we need to consider in which position the overlap happens in calculating P2? I thought order does not matter
Thank you.

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