To find the probability of obtaining \( k = 14 \) successes in \( n = 20 \) trials with a success probability of \( q = 0.25 \) on a single trial, we can use the binomial probability formula:
\[ P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \]
In this case:
- \( n = 20 \)
- \( k = 14 \)
- \( p = 0.25 \) (probability of success)
- \( 1 - p = 0.75 \) (probability of failure)
First, calculate \( \binom{20}{14} \):
\[ \binom{20}{14} = \binom{20}{6} = \frac{20!}{14! \cdot 6!} = \frac{20 \times 19 \times 18 \times 17 \times 16 \times 15}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 38760 \]
Next, we compute \( p^k \) and \( (1 - p)^{n - k} \):
- \( p^k = 0.25^{14} \)
- \( (1 - p)^{n - k} = 0.75^{6} \)
Now calculate these values:
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Calculate \( 0.25^{14} \): \[ 0.25^{14} = \left(\frac{1}{4}\right)^{14} = \frac{1}{4^{14}} = \frac{1}{268435456} \approx 3.725 \times 10^{-9} \]
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Calculate \( 0.75^{6} \): \[ 0.75^6 = 0.17803125 \]
Now, substitute these values into the binomial formula:
\[ P(X = 14) = \binom{20}{14} (0.25)^{14} (0.75)^{6} \] \[ P(X = 14) = 38760 \cdot (3.725 \times 10^{-9}) \cdot (0.17803125) \]
Now, perform the multiplication:
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Calculate \( 38760 \cdot 3.725 \times 10^{-9} \): \[ 38760 \cdot 3.725 \approx 144.0795 \times 10^{-9} = 1.440795 \times 10^{-7} \]
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Finally, multiply by \( 0.17803125 \): \[ P(X = 14) \approx 1.440795 \times 10^{-7} \cdot 0.17803125 \approx 2.56619692 \times 10^{-8} \]
Thus, rounding to four decimal places,
\[ P(X = 14) \approx 0.0000 \quad \text{(since the value is extremely small)} \]
So the probability of getting exactly 14 successes in this scenario is very close to 0. Thus, the correct answer is:
\[ P(X = 14) \approx 0.0000 \]
This indicates the likelihood of achieving such a high number of successes (14) with a relatively low probability of success (25%) is effectively negligible.
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