The probability that a non leap year will have 53 Thursdays is
- \(\frac{1}{221}\)
- \(\frac{1}{7}\)
- \(\frac{6}{7}\)
- \(\frac{9}{13}\)
The Correct Option is B
Solution and Explanation
Correct answer: \(\frac{1}{7}\)
Explanation:
In a non-leap year, there are 365 days. Dividing 365 by 7 gives: \[ 365 \div 7 = 52 \text{ weeks and } 1 \text{ extra day} \] Therefore, there will be exactly 52 occurrences of each day of the week, plus one additional day. The extra day could be any of the seven days of the week, so there is a \( \frac{1}{7} \) chance that the extra day will be a Thursday. If the extra day is a Thursday, there will be 53 Thursdays in that year. Hence, the probability that a non-leap year will have 53 Thursdays is \( \frac{1}{7} \).
Therefore, the probability is \(\frac{1}{7}\).
Top Questions on Probability
Based upon the results of regular medical check-ups in a hospital, it was found that out of 1000 people, 700 were very healthy, 200 maintained average health and 100 had a poor health record.
Let \( A_1 \): People with good health,
\( A_2 \): People with average health,
and \( A_3 \): People with poor health.
During a pandemic, the data expressed that the chances of people contracting the disease from category \( A_1, A_2 \) and \( A_3 \) are 25%, 35% and 50%, respectively.
Based upon the above information, answer the following questions:
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