Question

Thirty days are in September, April, June and November; the other months have thirty-one days (except February). A month is chosen at random. What is the probability that the chosen month has exactly three days less than the maximum of 31, that is, exactly 28 days?

• A. \(\frac{15}{16}\)
• B. 1
• C. \(\frac{3}{48}\)
• D. None of these

Hint:

Three days less than 31 is 28 days; only February has 28 days, so the probability is \(\frac{1}{12}\). Check whether this matches any listed fraction.

Answer 1

The Correct Option is D

Step 1: Work out the target number of days.
The maximum number of days in any month is 31. Three days less than this maximum is \(31 - 3 = 28\) days.

Step 2: Find which months have exactly 28 days.
Out of the 12 months, only February has 28 days on a standard (non-leap) year, which is the usual assumption in this kind of question.

Step 3: Set up the probability.
Probability is favourable outcomes divided by total outcomes. There are 12 months in total and only 1 of them (February) has 28 days.
\[ P(\text{28 days}) = \frac{1}{12} \]

Step 4: Compare this with the given options.
Option A is \(\frac{15}{16} = 0.9375\), far too large to be a 1-in-12 chance.
Option B is 1, meaning certainty, which is clearly wrong since only 1 of 12 months qualifies.
Option C is \(\frac{3}{48} = \frac{1}{16} = 0.0625\), close to but not equal to \(\frac{1}{12} \approx 0.0833\).
Since \(\frac{1}{12}\) does not match any of A, B or C, the correct choice must be "None of these".

Final Answer:
The probability is \(\frac{1}{12}\), which is not listed among the first three options.
\[ \boxed{\text{None of these}} \]