Question 293891
A state lottery requires you to pick 6 different numbers from 1 to 40 to win $1,000,000. A) what is the probability of winning if order is not important?
<pre><font size = 4 color = "indigo"><b> 
Calculate the numerator:
There is only one winning combination.  So the numerator of the 
desired probability is 1.

Calculate the denominator:
There are "40 numbers, choose 6", or 40C6 possible combinations.
So the denominator of the desired probability is 40C6.

So the probability is {{{1/(40C6)=1/3838380}}}
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B) If you pick 5 of the 6 numbers correctly you win $10,000. What is the probabilty of winning?
<pre><font size = 4 color = "indigo"><b>
Suppose the winning 6 numbers were, say,

   6,  13,  26,  28,  31,  35

Calculate the numerator:

There are "6 numbers, choose 5" or 6C5 ways to pick 5 correctly. 
AND
you must also pick 1 wrong number.  That wrong number has to be
different from any of the 6 correct numbers, and so there are
40-6 or 34 wrong numbers you could have picked.  That's 34 wrong
numbers, choose 1 or 34C1. Since AND means to multiply, the
numerator is {{{6C5*34C1=6*34=204}}}

The denominator is the same 3838380 as in part (A)

So the desired probability is {{{204/3838380=17/319865}}}

However that's just the probability of winning the $10000.

The words "probability of winning" could be taken as the
probability of winning either $1,000,000 OR $10,000. So

P(winning $1,000,000 OR winning $10,000) and since 
"OR" means "ADD", we add the probability found in part A,

so the correct answer is {{{1/3838380+204/3838380=205/3838380=41/767676)}}},
the probability of winning a million or ten thousand.

Edwin</pre>