Question:
During the winter months in a certain village in Scotland, the probability of a day having severe fog is \(0.6\). The probability that in a given week there will be exactly two days with severe fog is:
During the winter months in a certain village in Scotland, the probability of a day having severe fog is \(0.6\). The probability that in a given week there will be exactly two days with severe fog is:
Show Hint
Whenever a question asks for exactly \(r\) successes out of \(n\) independent trials, use the binomial formula:
\[
P(X=r)=\,{}^{n}C_r p^r(1-p)^{n-r}.
\]
Updated On: Jun 26, 2026
- \(\dfrac{6048}{5^7}\)
- \(\dfrac{2016}{5^7}\)
- \(\dfrac{3024}{5^7}\)
- \(\dfrac{12096}{5^7}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Identify the probability distribution.
Each day can be regarded as a Bernoulli trial.
Probability of severe fog on a day: \[ p=0.6=\frac{3}{5} \] Probability of no severe fog on a day: \[ q=1-p=0.4=\frac{2}{5} \] A week consists of \(7\) days.
We need the probability of exactly \(2\) days having severe fog.
Step 2: Use the Binomial Probability Formula.
For exactly \(r\) successes in \(n\) trials, \[ P(X=r) = {}^{n}C_{r}p^{r}q^{\,n-r} \] Here, \[ n=7,\qquad r=2 \] Hence, \[ P(X=2) = {}^{7}C_{2} \left(\frac{3}{5}\right)^2 \left(\frac{2}{5}\right)^5 \]
Step 3: Simplify the expression.
Since \[ {}^{7}C_{2} = 21 \] we get \[ P(X=2) = 21\cdot \frac{9}{25} \cdot \frac{32}{3125} \] \[ = \frac{21\cdot 9\cdot 32}{5^7} \] \[ = \frac{6048}{5^7} \]
Step 4: Final conclusion.
Therefore, \[ \boxed{\frac{6048}{5^7}} \] Hence, the correct option is \[ \boxed{(1)\ \frac{6048}{5^7}} \]
Each day can be regarded as a Bernoulli trial.
Probability of severe fog on a day: \[ p=0.6=\frac{3}{5} \] Probability of no severe fog on a day: \[ q=1-p=0.4=\frac{2}{5} \] A week consists of \(7\) days.
We need the probability of exactly \(2\) days having severe fog.
Step 2: Use the Binomial Probability Formula.
For exactly \(r\) successes in \(n\) trials, \[ P(X=r) = {}^{n}C_{r}p^{r}q^{\,n-r} \] Here, \[ n=7,\qquad r=2 \] Hence, \[ P(X=2) = {}^{7}C_{2} \left(\frac{3}{5}\right)^2 \left(\frac{2}{5}\right)^5 \]
Step 3: Simplify the expression.
Since \[ {}^{7}C_{2} = 21 \] we get \[ P(X=2) = 21\cdot \frac{9}{25} \cdot \frac{32}{3125} \] \[ = \frac{21\cdot 9\cdot 32}{5^7} \] \[ = \frac{6048}{5^7} \]
Step 4: Final conclusion.
Therefore, \[ \boxed{\frac{6048}{5^7}} \] Hence, the correct option is \[ \boxed{(1)\ \frac{6048}{5^7}} \]
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