Find the probability of getting 53 Fridays in a leap year.a)3/7b)4/7c)...
To find the probability of getting 53 Fridays in a leap year, we need to consider the total number of possible outcomes and the number of favorable outcomes.
Total number of outcomes:
In a leap year, there are 366 days.
Number of favorable outcomes:
To have 53 Fridays in a leap year, we need to have 53 days that fall on a Friday. There are 7 possible days of the week, so we can choose any 53 days out of the 366 days to be Fridays.
Using the combination formula, the number of ways to choose 53 days out of 366 is given by:
C(366, 53) = 366! / (53! * (366-53)!)
= (366 * 365 * 364 * ... * 315) / (53 * 52 * 51 * ... * 3 * 2 * 1)
Simplifying this expression would give us a very large number, but we don't need the exact value. We only need the probability, which is the ratio of the number of favorable outcomes to the total number of outcomes.
Calculating the probability:
The probability can be calculated as:
P = Number of favorable outcomes / Total number of outcomes
Since we have already determined that the total number of outcomes is 366 and the number of favorable outcomes is C(366, 53), we can substitute these values into the formula:
P = C(366, 53) / 366
Calculating this expression would give us a decimal value. To convert it into a fraction, we can simplify the expression by dividing both the numerator and denominator by their greatest common divisor.
After simplification, the probability can be written as a fraction, and the corresponding option is the correct answer.
Find the probability of getting 53 Fridays in a leap year.a)3/7b)4/7c)...
Leap year contains 366 days.
⇒ 52 weeks + 2 days
52 weeks contain 52 Fridays.
We will get 53 Fridays if one of the remaining two days is a Friday. Total possibilities for two days are:
(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)
There are 7 possibilities and out of these 2 are favourable cases.
Required probability = 2/7