C(66;25) / C(70;25) How would I solve this? ^
I just need it interpreted mainly
@Luis_Rivera @Mertsj
Is that 66 choose 25 divided by 70 choose 25 or what?
it has to do with probability
the semicolon can be a comma too
\[\frac{66!}{25!(66-25)!}\]
Evaluate that and divide it by this: \[\frac{70!}{25!(70-25!)}\]
im getting an error for the 2nd one
got 2.933706409 for the first
my calc says: work overflow
\[\frac{\frac{66!}{25!41!}}{\frac{70!}{25!45!}}=\frac{66!}{25!41!}\times \frac{25!45!}{70!}=\]
\[\frac{66!45\times44\times43\times42\times41!}{41!70\times69\times68\times67\times66!}=\]
\[\frac{45\times44\times 43\times 42}{70\times69\times68\times67}\]
Perhaps you can finish it.
What is the probability that at least 2 people in a random group of 6 people have a birthday on the same day of the week?
can you help me with this question too?
the answer is p(7,6)/(7^6)=.0428 1-.0428=9572.... but i dont know how to put the numerator into a calc.
@Mertsj
p(7,6)=7!/(7-6)!
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