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You flip a coin 10 times. Knowing that the event satisfies the requirements for a binomial distribution, find the probability that exactly 7 of the outcomes are heads.

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Assuming a fair coin, and requirements for a binormial distribution are satisfied, then 
p=0.50
n=20
The number of successes, x, is then given by[tex]P(x)=C(n,x)p^x(1-p)^{n-x}[/tex]where[tex]C(n,x)=\frac{n!}{x!(n-x)!}[/tex]
For x=7,
[tex]P(x)=C(n,x)p^x(1-p)^{n-x}[/tex]
[tex]P(7)=C(n,x)p^x(1-p)^{n-x}[/tex]
[tex]=C(20,7)0.5^7(0.5)^{20-7}[/tex]
[tex]=C(20,7)0.5^{20}[/tex]
[tex]=\frac{77520}{1048576}[/tex]
[tex]=\frac{4845}{65536}[/tex]
[tex]=0.07393[/tex]    to 5 places of decimal

Ans. the probability is 0.07393 to 5 places of decimal.

Answer:

0.1172

Step-by-step explanation:

The formula for the probability in a binomial distribution (n trials, r successes) is

[tex]\binom{n}{r}(p)^r(1-p)^{n-r}[/tex] where p is the probability of success.

For flipping a coin, the probability of success, p, is 0.5.

The number of trials, n, is 10.  The number of successes, r, is 7.  This gives us

[tex]\binom{10}{7}(0.5)^7(1-0.5)^{10-7}\\\\\binom{10}{7}(0.5)^7(0.5)^3=0.1171875 \approx 0.1172[/tex]

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