Aron flips a penny 9 times. Which expression represents the probability of getting exactly 3 heads?

[tex]\[
\begin{array}{c}
P(k \text { successes })={ }_n C_k p^k(1-p)^{n-k} \\
{ }_n C _k=\frac{n!}{(n-k)!k!}
\end{array}
\][/tex]

[tex]\[
\begin{array}{l}
A. { }_9 C_3(0.5)^3(0.5)^6 \\
B. { }_9 C_3(0.5)^3 \\
C. { }_9 C_3(0.5)^3(0.5)^9 \\
D. { }_9 C_6(0.5)^6
\end{array}
\][/tex]



Answer :

To determine the probability of getting exactly 3 heads out of 9 flips of a penny, we can use the binomial probability formula. The formula is given by:

[tex]\[ P(k \text{ successes}) = \binom{n}{k} p^k (1-p)^{n-k} \][/tex]

where:
- [tex]\( n \)[/tex] is the number of trials (flips), which is 9 in this case.
- [tex]\( k \)[/tex] is the number of successes (heads), which is 3 in this case.
- [tex]\( p \)[/tex] is the probability of success (getting a head) in each trial, which is 0.5 for flipping a fair penny.

The binomial coefficient [tex]\(\binom{n}{k}\)[/tex] is calculated by:

[tex]\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \][/tex]

Plugging in the values:
- [tex]\( n = 9 \)[/tex]
- [tex]\( k = 3 \)[/tex]
- [tex]\( p = 0.5 \)[/tex]

The expression becomes:

[tex]\[ P(3 \text{ heads}) = \binom{9}{3} (0.5)^3 (1-0.5)^{9-3} \][/tex]

Simplifying further:

[tex]\[ P(3 \text{ heads}) = \binom{9}{3} (0.5)^3 (0.5)^6 \][/tex]

Combining the powers of 0.5:

[tex]\[ P(3 \text{ heads}) = \binom{9}{3} (0.5)^{3+6} \][/tex]
[tex]\[ P(3 \text{ heads}) = \binom{9}{3} (0.5)^9 \][/tex]

The correct expression among the given choices is:

[tex]\[ { }_9 C_3(0.5)^3(0.5)^6 \][/tex]

This simplifies to:

[tex]\[ { }_9 C_3(0.5)^9 \][/tex]

Hence, the correct expression that represents the probability of getting exactly 3 heads is:

[tex]\[ { }_9 C_3(0.5)^3(0.5)^6 \][/tex]

The numerical outcome for the given problem would be [tex]\( 84.0 \)[/tex] for the combination calculation and [tex]\( 0.1640625 \)[/tex] for the probability.