statistics, probability, birthday problem, person selected at random, his or her birthday is...?

Question

I am stuck on another problem as I am studying.

I would appreciate the help. I dont ike the birthday problem. I know its out of 365 days...and I get some of it...to a point.

IV. If a person is selected at random, find the probability that his or her birthday is
a) October 8
b) in October
c) on a Wednesday
d) on a day of the week starts with T.

Answer 1

a) 1/365
b) 31/365
c) 1/7
d) 2/7

For C and D, yes there are 52 weeks in a year. But in a year with 365 days that means 52 weeks plus one extra day. So I don't think you can say the chance of a Wednesday is 52/365 without knowing what the extra day is. Over the long run, without knowing what kind of year it is, the chance of drawing a Wednesday will still be 1/7 and a day with a T is 2/7, not 52/365 or 104/365.

Answer 2

probability is defined as the number of possible success divided by the total number of possible outcomes.

there are 365 possible days in the year to have a birthday (forgetting about leap years)

there is only one October 8 in the year so you have a 1/365 probability of the person's birthday being that day.

the month of october has 31 days so there is a 31/365 probability of having a birthday in october

there are 52 weeks in a year, so 52 wednesdays, 52/365 for a birthday on a wednesday

two days out of the week start with a T so 104/365 for a birthday on a day of the week starting with T.

Answer 3

Well, as you said, there are 365 days in the year. So, October 8 is 1 day. So, the odds are 1/365.

The month of October is 31 days...so that answer is 31/365. That might simplify (just pretend it's a fraction and simplify it.)

Wednesday...there is 1 wednesday per week. 52 weeks in a year, so 52 wednesdays in a year. 52/365

Two days of the week start with T, Tuesday and Thursday, same thing that we did with wednesday, only twice as many.

104/365

Be sure to simplify these if they do!

Answer 4

All of these are so many chances in 365, so you just need to find the first number in every case

There is only one October 8th in the year, so this is 1 in 365
There are 31 days in October, so this one is 31 in 365
There are 52 Wednesdays in a year, so this is 52 in 365
There are two days in every week that start with T and 52 weeks in the year, so this has the greatest chance of happening at 104 in 365

Answer 5

a) on October 8
since there is only one day in a year that is Oct. 8
so P = 1/365

b) in October
there are 31 days in October
P = 31/365

c) on Wednesday
there is only day that is Wednesday in on 1 week. We know that there are 52 weeks and 1 day in one year. Since i takes 7 more days to be back on wednesday again, the 1 day does not count. So there are 52 wednesday in 1 week

P = 52/365


d) on the day of the week that starts with T
Tuesday, thursday.

there are 104 days that start with T
so P = 104/365

Answer 6

Everything said here will end with "out of 365".

a) How many October 8ths are there in a year?
b) How many October days are there in a year?
c) How many Wednesdays are there in a year?
d) How many days of the week begin with T and then how many weeks are there in a year? Multiply the two numbers together.