Probability question. I'm asking for the best method to solve it. I don't need the numerical answer.?

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This is a binomial distribution. Since n is so large the binomial can be cumbersome to compute. So for large n use the normal approximation to the binomial

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Do I need to calculate the prob of winning exactly 71 games, then prob of winning 72 games, and so on up to 100, then add those probabilities together? Is there a better faster way to do it?
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You final answer will be more accurate if you computed for each exact value but this tends to be too cumbersome and time consuming.

n = 100 
p = ⅗

µ = np = 100( ⅗ ) = 60
σ² = np( 1 - p ) = 60( 1 - ⅗ ) = 24
∴σ = √24

…… winning more than 70 games implies using x > 70.5 for normal approximation to the binomial

z = ( x - µ ) / σ
z = ( 70.5 - 60 ) / √24
z = 2.1433

from the table: 
z = 2.14, P = 0.9838
z = 2.15, P = 0.9842

by interpolation, 
(2.1433 - 2.14)/(P - 0.9838) = (2.15-2.1433)/(0.9842 - P)
P ≈ 0.9839

P( winning > 70 ) = 1 - P( @z = 2.1433 )
P( winning > 70 ) = 1 - 0.9839
P( winning > 70 ) = 0.0161 = 1.61%
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You outlined how to get the exact answer. But with n = 100, you can, and are probably supposed to, use the normal distribution to the binomial.

The properties of the binomial are well-known. The mean is np = 100 x 3/5 = 60
The variance is npq = 100(3/5)(2/5) = 24. So the s.d. is √24 = 4.899.

Now you just use those facts and pretend you have a normal distribution.
the answer is the area under the "normal" curve from 70.5 to infinity.
You really want 71 and above, but when you use this approximation you extend things by 1/2, so 71 is represented by the interval (70.5, 71.5).

So 70.5 is 70.5 - 60 = 10.5 above the mean.
That is 10.5 / 4.899 standard deviations.

Now just calculate that and use a Z-table to get the area you want.

See https://www.youtube.com/watch?v=CCqWkJ_pqNU
for a tutorial....Show more