Question 76116
<pre><font size = 5><b>
A coin is tossed 9 times. What is the 
probability that the coin will land heads 
at least 7 times?

This is a binomial probability. You can 
find it directly from a binomial
table, a TI-83 calculator, or entirely 
from the formula.

Using the formula: 

{{{C(n,x)p^n*q^(n-x)}}}

gives the probability that in n trials 
there will be exactly x successes
where

p = the probability of one success in 
one trial and q = 1-p = the probability 
of one failure in one trial.

{{{C(n,x) = n!/(x!(n-x)!)}}}

In the case of a coin, p = 1/2 and q = 1 - 1/2 = 1/2

P(coin landing heads at least 7 times out of 9 tosses) =

P(coin landing exactly 7 times out of 9) +   
       P(coin landing exactly 8 times out of 9) +
              P(coin landing exactly 9 times out of 9) =

{{{C(9,7)(1/2)^7(1/2)^2 + C(9,8)(1/2)^8(1/2)^1 + C(9,9)(1/2)^9(1/2)^0}}} =

{{{C(9,7)(1/2)^9 + C(9,8)(1/2)^9 + C(9,9)(1/2)^9}}} =

Factor out {{{(1/2)^9}}}

{{{(1/2)^9*(C(9,7) + C(9,8) + C(9,9))}}} =

{{{(1/512)*(C(9,7) + C(9,8) + C(9,9))}}}

which gives the desired probability.

To calculate this we need:

{{{C(9,7) = 9!/(7!(9-7)!)= 9!/(7!2!)=(9*8*7*6*5*4*3*2*1)/((7*6*5*4*3*2*1)(2*1))=36}}}

{{{C(9,8) = 9!/(8!(9-8)!)= 9!/(8!1!)=(9*8*7*6*5*4*3*2*1)/((8*7*6*5*4*3*2*1)(1))=9}}}

{{{C(9,9) = 9!/(9!(9-9)!)= 9!/(9!0!)=(9*8*7*6*5*4*3*2*1)/((9*7*6*5*4*3*2*1)(1))=1}}}

Therefore the desired probability =

{{{(1/512)*(C(9,7) + C(9,8) + C(9,9)) = (1/512)(36+9+1)= 46/512 = 23/256 = .08984375}}} 

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If you use a TI-83 or TI-84 calculator, type in

1 - binomcdf(9,1/2,6)

then press ENTER and you get .08984375

To get binomcdf( , press 2nd, then VARS, then 
scroll down to binomcdf( , then press ENTER.

Edwin</pre>