Question

A day can only be cloudy or sunny. The probability of a day being cloudy is 0.5, independent of the condition on other days. What is the probability that in any given four days, there will be three cloudy days and one sunny day?

• A. \(\dfrac{1}{4}\)
• B. \(\dfrac{3}{4}\)
• C. \(\dfrac{2}{3}\)
• D. \(\dfrac{3}{8}\)

Hint:

Count how many of the four days can be the single sunny day, then multiply by the chance of one fixed order of three cloudy and one sunny day.

Answer 1

The Correct Option is A

Step 1: Set up the probability model.
Each day is cloudy with probability \(0.5\) and sunny with probability \(0.5\), and the four days are independent of each other. This is a binomial setting with \(n=4\) trials and success probability \(p=0.5\), where a "success" is a cloudy day.

Step 2: Write the binomial probability formula.
The probability of getting exactly \(k\) cloudy days out of \(n\) days is
\[ P(k) = \binom{n}{k}p^k(1-p)^{n-k} \]
Here we want exactly \(3\) cloudy days out of \(4\), so \(n=4\), \(k=3\), \(p=0.5\).

Step 3: Count the number of ways to arrange three cloudy days.
\[ \binom{4}{3}=4 \]
This counts the four ways to choose which one of the four days is the sunny day, since the other three are then cloudy.

Step 4: Substitute into the formula.
\[ P(3) = \binom{4}{3}(0.5)^3(0.5)^1 = 4\times(0.5)^4 \]
\[ (0.5)^4 = \frac{1}{16} \]
So
\[ P(3) = 4\times\frac{1}{16} = \frac{4}{16} = \frac{1}{4} \]

Step 5: Rule out the other options.
Option (B) \(\dfrac{3}{4}\) is close to the complement of the answer and does not follow from the binomial formula for exactly three cloudy days.
Option (C) \(\dfrac{2}{3}\) does not arise from any valid substitution into \(\binom{4}{3}(0.5)^4\), it is simply a distractor value.
Option (D) \(\dfrac{3}{8}\) would result from forgetting to raise \(0.5\) to the full fourth power or mis-counting the arrangements, so it is also incorrect.

Step 6: Final Answer.
The probability of exactly three cloudy days and one sunny day in four days is
\[ \boxed{\dfrac{1}{4}} \]