(A) Let $p$ be the probability of severe fog on a day,so $p = 0.6 = \frac{3}{5}$.Let $q$ be the probability of no severe fog on a day,so $q = 1 - 0.6

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(A) Let $p$ be the probability of severe fog on a day,so $p = 0.6 = \frac{3}{5}$.Let $q$ be the probability of no severe fog on a day,so $q = 1 - 0.6 = 0.4 = \frac{2}{5}$.The number of days in a week is $n = 7$.Using the binomial distribution formula $P(X = k) = {n \choose k} p^k q^{n-k}$,for $k = 2$:$P(X = 2) = {7 \choose 2} \times (\frac{3}{5})^2 \times (\frac{2}{5})^5$$P(X = 2) = 21 \times \frac{9}{25} \times \frac{32}{3125}$$P(X = 2) = \frac{21 \times 9 \times 32}{5^7} = \frac{6048}{5^7}$.

Class 12 Mathematics Probability Binomial distribution

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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:

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A
$\frac{6048}{5^7}$
B
$\frac{2016}{5^7}$
C
$\frac{3024}{5^7}$
D
$\frac{12096}{5^7}$

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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:

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