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.
- 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,
\[ \binom{20}{14} = \frac{27907200}{720} = 38760 \]
- Calculate the probabilities: \[ p^k = (0.25)^{14} \] \[ q^{n-k} = (0.75)^{6} \]
Calculating each:
\[ (0.25)^{14} = 0.25^{14} = 0.00000005960464453125 \text{ (approximately)} \] \[ (0.75)^{6} = 0.75 \times 0.75 \times 0.75 \times 0.75 \times 0.75 \times 0.75 = 0.177978515625 \]
- 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:
\[ P(X = k) \approx 0.0004 \]