What is the probability of getting 53 Fridays in a non-leap year?with explanation pls
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according to me there is 0 probability of getting 53 fridays in a non leap year because there is maximum 4×12 Friday's in a year so there is no chance of 53 firdays in a year
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step 1 Possible events for 1 odd day
The odd day may be either Sunday, Monday, Tuesday, Wednesday, Thursday, Friday or Saturday. Therefore, the total number of possible outcome or elements of sample space is 7.
step 2 Probability of 1 Odd day to be Friday :
The sample space S = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
Expected event of A = {Friday}
P(A) =
{Friday}{Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
P(A) = 1/7
P(A) = 0.14
0.14 or 1/7 is probability for 53 Fridays in a non-leap year.
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