You have a probability of 3/5 of winning a game of chance. You play the

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mes. What is the prob of winning more than 70 games?

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?

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

oximation to the binomial

“

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?

“

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 –

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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Anonymous

Anonymous

ity.

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.