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
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.