Question

Assume that a procedure yields a binomial distribution with a trial repeated n=20
times. Use either the binomial probability formula (or technology) to find the probability of k=14
successes given the probability q=0.25 of success on a single trial.

(Report answer accurate to 4 decimal places.)

Answer the formula
P(X=k)=

the answer is not 0.0000

Answer 1

To calculate the probability of getting exactly \( k = 14 \) successes in \( n = 20 \) trials, where the probability of success on a single trial is \( p = 0.25 \) (and therefore the probability of failure \( q = 1 - p = 0.75 \)), we can use the binomial probability formula:

\[
P(X = k) = \binom{n}{k} p^k q^{n-k}
\]

Here, \( \binom{n}{k} \) is the binomial coefficient calculated as:

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

In our case:
- \( n = 20 \)
- \( k = 14 \)
- \( p = 0.25 \)
- \( q = 0.75 \)

Let's calculate it step-by-step.

1. **Calculate the binomial coefficient:**

\[
\binom{20}{14} = \frac{20!}{14!6!}
\]
Calculating \( 20! \), \( 14! \), and \( 6! \):
- \( 20! = 20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14! \)
- \( 14! = 14! \) (cancels out)
- \( 6! = 720 \)

So,

\[
\binom{20}{14} = \frac{20 \times 19 \times 18 \times 17 \times 16 \times 15}{6!} = \frac{20 \times 19 \times 18 \times 17 \times 16 \times 15}{720}
\]

Calculating this product:

\[
20 \times 19 = 380
\]
\[
380 \times 18 = 6840
\]
\[
6840 \times 17 = 116280
\]
\[
116280 \times 16 = 1860480
\]
\[
1860480 \times 15 = 27907200
\]
Thus,

\[
\binom{20}{14} = \frac{27907200}{720} = 38760
\]

2. **Calculate the probabilities:**
\[
p^k = (0.25)^{14}
\]
\[
q^{n-k} = (0.75)^{6}
\]

Calculating each:

\[
(0.25)^{14} = 0.25^{14} = 0.00000005960464453125 \text{ (approximately)}
\]
\[
(0.75)^{6} = 0.75 \times 0.75 \times 0.75 \times 0.75 \times 0.75 \times 0.75 = 0.177978515625
\]

3. **Putting it all together:**
\[
P(X = 14) = 38760 \times (0.25)^{14} \times (0.75)^{6}
\]
Substituting the values:

\[
P(X = 14) \approx 38760 \times 0.00000005960464453125 \times 0.177978515625
\]
Calculating this out:

\[
\approx 38760 \times 0.00000005960464453125 \times 0.177978515625 = 0.000408
\]

So, the probability of getting exactly 14 successes is

\[
P(X = 14) \approx 0.0004 \text{ (to four decimal places)}
\]

Therefore, the answer is:

\[
P(X = k) \approx 0.0004
\]