+870 A man plays the lottery, and wants to determinate his chances of winning. Six b***s are picked randomly out of ninety ones. Each number has a number on it (the numbers range from 1 to 90), and this number is the only difference between the b***s. The player can bet on 10 numbers. He gets £100,000 for each correct number (correct means that the player had bet on this number and the number is on one of the b***s which were picked out) plus 400,000 extra pounds if he had bet on all six numbers that were picked out.
What is the probability he gets:
a)£100,000 (1 correct number)
b)£200,000 (2 correct numbers)
c)£400,000 (4 correct numbers)
d)£500,000 (5 correct numbers)
e)£1,000,000 (6 correct numbers)
+118677 Hi Einstein Jr.
Here is how it is done 
The number of ways to choose 10 b***s from 90, 6 are drawn
So there are 84 Lose numbers and 6 Win numbers
It is easier if the winning numbers numbers are drawn first and and kept secret.
Then the player has to draw 10 ball numbers to match the winning ones (hopefully).
There are 90 numbers that can be drawn from so there are 90C10 possible draws
a) How many ways can just one be correct?
you want 9 fails and 1 success = 86C9*6C1
P(1 success)=84C9*6C1/90C10
$${\frac{{\left({\frac{{\mathtt{84}}{!}}{{\mathtt{9}}{!}{\mathtt{\,\times\,}}({\mathtt{84}}{\mathtt{\,-\,}}{\mathtt{9}}){!}}}\right)}{\mathtt{\,\times\,}}{\mathtt{6}}}{{\left({\frac{{\mathtt{90}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}} = {\mathtt{0.386\: \!113\: \!895\: \!203\: \!522\: \!6}}$$
b) P(2 successes)
84C8*6C2/90C10
$${\frac{{\left({\frac{{\mathtt{84}}{!}}{{\mathtt{8}}{!}{\mathtt{\,\times\,}}({\mathtt{84}}{\mathtt{\,-\,}}{\mathtt{8}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{6}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)}}{{\left({\frac{{\mathtt{90}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}} = {\mathtt{0.114\: \!310\: \!034\: \!764\: \!200\: \!8}}$$
c) P(3 successes)
84C7*6C3/90C10
$${\frac{{\left({\frac{{\mathtt{84}}{!}}{{\mathtt{7}}{!}{\mathtt{\,\times\,}}({\mathtt{84}}{\mathtt{\,-\,}}{\mathtt{7}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{6}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}}{{\left({\frac{{\mathtt{90}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}} = {\mathtt{0.015\: \!835\: \!156\: \!330\: \!971\: \!5}}$$
c) P(3 successes)
84C7*6C3/90C10
$${\frac{{\left({\frac{{\mathtt{84}}{!}}{{\mathtt{7}}{!}{\mathtt{\,\times\,}}({\mathtt{84}}{\mathtt{\,-\,}}{\mathtt{7}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{6}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}}{{\left({\frac{{\mathtt{90}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}} = {\mathtt{0.015\: \!835\: \!156\: \!330\: \!971\: \!5}}$$
d) P(4 successes)
84C6*6C4/90C10
$${\frac{{\left({\frac{{\mathtt{84}}{!}}{{\mathtt{6}}{!}{\mathtt{\,\times\,}}({\mathtt{84}}{\mathtt{\,-\,}}{\mathtt{6}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{6}}{!}}{{\mathtt{4}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{4}}){!}}}\right)}}{{\left({\frac{{\mathtt{90}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}} = {\mathtt{0.001\: \!065\: \!827\: \!829\: \!969\: \!2}}$$
e) P(5 successes)
84C5*6C5/90C10
$${\frac{{\left({\frac{{\mathtt{84}}{!}}{{\mathtt{5}}{!}{\mathtt{\,\times\,}}({\mathtt{84}}{\mathtt{\,-\,}}{\mathtt{5}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{6}}{!}}{{\mathtt{5}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{5}}){!}}}\right)}}{{\left({\frac{{\mathtt{90}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}} = {\mathtt{0.000\: \!032\: \!379\: \!579\: \!644\: \!6}}$$
f) P(6 successes)
84C4*6C6/90C10
$${\frac{{\left({\frac{{\mathtt{84}}{!}}{{\mathtt{4}}{!}{\mathtt{\,\times\,}}({\mathtt{84}}{\mathtt{\,-\,}}{\mathtt{4}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{6}}{!}}{{\mathtt{6}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{6}}){!}}}\right)}}{{\left({\frac{{\mathtt{90}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}} = {\mathtt{0.000\: \!000\: \!337\: \!287\: \!288}}$$
NOW for good measure
P(no successes)
84C10/90C10
$${\frac{{\left({\frac{{\mathtt{84}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{84}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}{{\left({\frac{{\mathtt{90}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}} = {\mathtt{0.482\: \!642\: \!369\: \!004\: \!403\: \!3}}$$
Check
P(of any number of wins)
nCr(84,9)*6/nCr(90,10)+
nCr(84,8)*nCr(6,2)/nCr(90,10)+
nCr(84,7)*nCr(6,3)/nCr(90,10)+
nCr(84,6)*nCr(6,4)/nCr(90,10)+
nCr(84,5)*nCr(6,5)/nCr(90,10)+
nCr(84,4)*nCr(6,6)/nCr(90,10)+
nCr(84,10)/nCr(90,10)
$${\frac{{\left({\frac{{\mathtt{84}}{!}}{{\mathtt{9}}{!}{\mathtt{\,\times\,}}({\mathtt{84}}{\mathtt{\,-\,}}{\mathtt{9}}){!}}}\right)}{\mathtt{\,\times\,}}{\mathtt{6}}}{{\left({\frac{{\mathtt{90}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\left({\frac{{\mathtt{84}}{!}}{{\mathtt{8}}{!}{\mathtt{\,\times\,}}({\mathtt{84}}{\mathtt{\,-\,}}{\mathtt{8}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{6}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)}}{{\left({\frac{{\mathtt{90}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\left({\frac{{\mathtt{84}}{!}}{{\mathtt{7}}{!}{\mathtt{\,\times\,}}({\mathtt{84}}{\mathtt{\,-\,}}{\mathtt{7}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{6}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}}{{\left({\frac{{\mathtt{90}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\left({\frac{{\mathtt{84}}{!}}{{\mathtt{6}}{!}{\mathtt{\,\times\,}}({\mathtt{84}}{\mathtt{\,-\,}}{\mathtt{6}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{6}}{!}}{{\mathtt{4}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{4}}){!}}}\right)}}{{\left({\frac{{\mathtt{90}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\left({\frac{{\mathtt{84}}{!}}{{\mathtt{5}}{!}{\mathtt{\,\times\,}}({\mathtt{84}}{\mathtt{\,-\,}}{\mathtt{5}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{6}}{!}}{{\mathtt{5}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{5}}){!}}}\right)}}{{\left({\frac{{\mathtt{90}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\left({\frac{{\mathtt{84}}{!}}{{\mathtt{4}}{!}{\mathtt{\,\times\,}}({\mathtt{84}}{\mathtt{\,-\,}}{\mathtt{4}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{6}}{!}}{{\mathtt{6}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{6}}){!}}}\right)}}{{\left({\frac{{\mathtt{90}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\left({\frac{{\mathtt{84}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{84}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}{{\left({\frac{{\mathtt{90}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)}}} = {\mathtt{1}}$$
