IF THREE PEOPLE ARE IN A GROUP,THE CHANCES THAT ANYY TWO HAVE THE SAME BIRTHDAY IS O.OO82 WHAT IS THE CHANCE THAT NO ONE IN THE GROUP HAS THE SAME BIRTHDAYAS ANY PERSON IN THE GROUP??
i guess 1-.0082
hard to read the all caps
.9918 is what i get
3 in a group; .0082 chance the 2 have same birthday
naw thats wrong its either 0.9918
nvm your right
there are 8 outcomes all together
whew!
thanks
@amistre doesn't matter how many outcomes. this just says the probability of A is .0083 so the probability of not A is 1-.0083
or is it 4 outcomes :) nnn nyy yny yyy
actually it is a bit more complicated i think
itd help better if I could actually read the question ;)
i mean the answer is given, but the calculation to get it is a bit more
i was wondering if binomial fit; and such ... but yeah
no
i was confusing at least 2; as well
but actually easy enough to calculate no two same birthdays. pick one. they were born some day. forgetting about leap years then the probability that the next person is not born on the same day is \[\frac{364}{365}\]
and the next not either is \[\frac{363}{365}\] and their product is .9918 to four decimal places
more interestingly continuing in this pattern you only need to get two about 23 for the probability to be less than a half, and at around 30 it is only about 30% meaning that in a group of 30 people there is a 70% chance that two or more share a birthday
side question: what is a t-distribution?
side answer. no idea, sorry
i remember something called "students t" but all i remember is the name
we are going over how to find zscores and area at teh moment and the later chapters build into chi^2 and t distribution thanx to a central limit thrm
degrees of freedom, hypot testing and confidence intervals ...
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