Question 888129
<pre>
Do you mean getting 7 AT LEAST twice or EXACTLY twice?

Either way we first get the probability of getting a seven 
with just ONE roll of the dice:
<b>
(1,1) (1,2) (1,3) (1,4) (1,5) <font color="red">(1,6)</font>

(2,1) (2,2) (2,3) (2,4) <font color="red">(2,5)</font> (2,6) 

(3,1) (3,2) (3,3) <font color="red">(3,4)</font> (3,5) (3,6) 

(4,1) (4,2) <font color="red">(4,3)</font> (4,4) (4,5) (4,6) 

(5,1) <font color="red">(5,2)</font> (5,3) (5,4) (5,5) (5,6) 

<font color="red">(6,1)</font> (6,2) (6,3) (6,4) (6,5) (6,6) 
</b>
That's 6 ways out of 36 or {{{6/36}}} or {{{1/6}}} as the probability
of getting seven out with just ONE roll.

To get exactly 2 seven rolls out of 4, we use the formula

{{{"P(n,x;p)"}}}{{{""=""}}}{{{expr(n!/(x!(n-x)!))*p^x*(1-p)^(n-x)}}}

with {{{n=4}}}, {{{x=2}}}, {{{p=1/6}}}

{{{"P(4,2;1/6)"}}}{{{""=""}}}{{{expr(4!/(2!(4-2)!))*(1/6)^2*(1-1/6)^(4-2)}}}{{{""=""}}}{{{25/216}}}{{{""=""}}}{{{0.1157407407}}}

That's the probability of getting exactly 2 sevens.

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To get the probability of 2 or more, calculate the probabilities of

exactly 3 and exactly 4 and add them to the above.

With a TI-83 or TI-84 calculator, binompdf( is for "exactly",
binomcdf( is for "exactly or fewer")

binompdf(4,1/6,2) under 2ND VARS

To get probability of 2 or more, 1 minus probability of 1 or fewer:

1-binomcdf(4,1/6,1)  get .1319444444 which is {{{19/144}}}

Edwin</pre>