To determine the probability of getting exactly 3 heads when Aron flips a penny 9 times, we need to use the binomial probability formula:
[tex]\[ P(k \text{ successes}) = {^n}C_k \cdot p^k \cdot (1-p)^{n-k} \][/tex]
where:
- [tex]\( n \)[/tex] is the number of trials (flips).
- [tex]\( k \)[/tex] is the number of successes (heads).
- [tex]\( p \)[/tex] is the probability of success on a single trial.
- [tex]\( {^n}C_k \)[/tex] is the binomial coefficient, which represents the number of ways to choose [tex]\( k \)[/tex] successes in [tex]\( n \)[/tex] trials.
Given the values:
- [tex]\( n = 9 \)[/tex] (number of flips)
- [tex]\( k = 3 \)[/tex] (number of heads we want)
- [tex]\( p = 0.5 \)[/tex] (probability of getting heads in a single flip)
The binomial coefficient [tex]\( {^9}C_3 \)[/tex] can be calculated as:
[tex]\[ {^9}C_3 = \frac{9!}{3!(9-3)!} \][/tex]
Next, we can substitute [tex]\( n \)[/tex], [tex]\( k \)[/tex], and [tex]\( p \)[/tex] into the probability formula:
[tex]\[ P(3 \text{ heads}) = {^9}C_3 \cdot (0.5)^3 \cdot (1-0.5)^{9-3} \][/tex]
Simplifying [tex]\( (1-0.5) \)[/tex] gives [tex]\( 0.5 \)[/tex], thus:
[tex]\[ P(3 \text{ heads}) = {^9}C_3 \cdot (0.5)^3 \cdot (0.5)^6 \][/tex]
[tex]\[ = {^9}C_3 \cdot (0.5)^{3+6} \][/tex]
[tex]\[ = {^9}C_3 \cdot (0.5)^9 \][/tex]
Now, identifying the expression we used:
[tex]\[ {^9}C_3 \cdot (0.5)^3 \cdot (0.5)^6 \][/tex]
This matches the first option given:
[tex]\[ {^9}C_3(0.5)^3(0.5)^6 \][/tex]
Therefore, the correct expression that represents the probability of getting exactly 3 heads in 9 flips is:
[tex]\[ {^9}C_3(0.5)^3(0.5)^6 \][/tex]
