To solve this problem, we need to compute the binomial probability [tex]\( P(x) \)[/tex] for [tex]\( x = 3 \)[/tex] successes in [tex]\( n = 10 \)[/tex] trials, where the probability of success in each trial is [tex]\( p = 0.4 \)[/tex].
The formula for the binomial probability is given by:
[tex]\[
P(x) = \binom{n}{x} p^x (1 - p)^{n - x}
\][/tex]
where:
- [tex]\(\binom{n}{x}\)[/tex] is the combination of [tex]\( n \)[/tex] items taken [tex]\( x \)[/tex] at a time, also written as "n choose x,"
- [tex]\( p \)[/tex] is the probability of success,
- [tex]\( 1 - p \)[/tex] is the probability of failure,
- [tex]\( n \)[/tex] is the total number of trials,
- [tex]\( x \)[/tex] is the number of successful trials.
Let's break this down step-by-step:
1. Calculate the combination [tex]\(\binom{10}{3}\)[/tex]:
[tex]\[
\binom{n}{x} = \frac{n!}{x!(n - x)!}
\][/tex]
For [tex]\( n = 10 \)[/tex] and [tex]\( x = 3 \)[/tex]:
[tex]\[
\binom{10}{3} = \frac{10!}{3!(10 - 3)!} = \frac{10!}{3! \cdot 7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120
\][/tex]
2. Calculate [tex]\( p^x \)[/tex]:
[tex]\[
0.4^3 = 0.4 \times 0.4 \times 0.4 = 0.064
\][/tex]
3. Calculate [tex]\( (1 - p)^{n - x} \)[/tex]:
[tex]\[
(1 - 0.4)^{10 - 3} = 0.6^7
\][/tex]
We know [tex]\( 0.6^7 \approx 0.0279936 \)[/tex].
4. Combine these values into the binomial probability formula:
[tex]\[
P(3) = \binom{10}{3} \times 0.4^3 \times 0.6^7
\][/tex]
Substitute the values we have:
[tex]\[
P(3) = 120 \times 0.064 \times 0.0279936
\][/tex]
Perform the multiplication:
[tex]\[
P(3) = 120 \times 0.001791616 = 0.21499084799999998
\][/tex]
Rounding this value to four decimal places:
[tex]\[
P(3) \approx 0.2150
\][/tex]
Therefore, the probability of 3 successes in 10 independent trials with a success probability of 0.4 is approximately [tex]\( 0.2150 \)[/tex].
