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

• A. \( \frac{6048}{5^7} \)
• B. \( \frac{2016}{5^7} \)
• C. \( \frac{3024}{5^7} \)
• D. \( \frac{12096}{5^7} \)

Hint:

Binomial Probability: \(P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\).
\(n\) = number of trials (days in a week).
\(k\) = number of successes (days with severe fog).
\(p\) = probability of success on a single trial.
\(\binom{n}{k} = \frac{n!}{k!(n-k)!}\).

Answer 1

The Correct Option is A

This is a binomial probability problem. Let \(n\) be the number of days in a week, so \(n=7\). Let \(p\) be the probability of a day having severe fog, \(p = 0.6 = 6/10 = 3/5\). Let \(q\) be the probability of a day NOT having severe fog, \(q = 1-p = 1 - 0.6 = 0.4 = 4/10 = 2/5\). We want to find the probability of having exactly \(k=2\) days with severe fog in a week. The binomial probability formula is \(P(X=k) = \binom{n}{k} p^k q^{n-k}\). Here, \(n=7, k=2, p=3/5, q=2/5\). \(P(X=2) = \binom{7}{2} (3/5)^2 (2/5)^{7-2}\) \(P(X=2) = \binom{7}{2} (3/5)^2 (2/5)^5\) Calculate \(\binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21\). \((3/5)^2 = 9/25\). \((2/5)^5 = 2^5 / 5^5 = 32 / 3125\). So, \(P(X=2) = 21 \times \frac{9}{25} \times \frac{32}{3125}\). The denominator is \(25 \times 3125 = 5^2 \times 5^5 = 5^7\). \(5^7 = 5^2 \cdot 5^5 = 25 \cdot 3125\). \(25 \times 3125 = 25 \times (3000 + 100 + 25) = 75000 + 2500 + 625 = 77500 + 625 = 78125\). Numerator = \(21 \times 9 \times 32\). \(21 \times 9 = 189\). \(189 \times 32\): 189 x 32 ----- 378 (189 * 2) 5670 (189 * 30) ----- 6048 So, \(P(X=2) = \frac{6048}{5^7} = \frac{6048}{78125}\). This matches option (a). \[ \boxed{\frac{6048}{5^7}} \]