Skip to main content

So I was messing around with it more and talking to a friend about it. Can anyone tell me if this is correct...

$P(W) = 1 - P(W')$

$ = 1 - \left(\frac{_{13}C_3}{_{15C_3}}\right)^3 $$ = 1 - \left(\frac{^{13}C_3}{^{15}C_3}\right)^3 $

Which comes out to $P(W) = 0.752$

It just seems quite high to me.

So I was messing around with it more and talking to a friend about it. Can anyone tell me if this is correct...

$P(W) = 1 - P(W')$

$ = 1 - \left(\frac{_{13}C_3}{_{15C_3}}\right)^3 $

Which comes out to $P(W) = 0.752$

It just seems quite high to me.

So I was messing around with it more and talking to a friend about it. Can anyone tell me if this is correct...

$P(W) = 1 - P(W')$

$ = 1 - \left(\frac{^{13}C_3}{^{15}C_3}\right)^3 $

Which comes out to $P(W) = 0.752$

It just seems quite high to me.

deleted 2 characters in body
Source Link

So I was messing around with it more and talking to a friend about it. Can anyone tell me if this is correct...

$P(W) = 1 - P(W')$ =

$ = 1 - \left(\frac{_{13}C_3}{_{15C_3}}\right)^3 $

Which comes out to $P(W) = 0.752$

It just seems quite high to me.

So I was messing around with it more and talking to a friend about it. Can anyone tell me if this is correct...

$P(W) = 1 - P(W')$ =

$ = 1 - \left(\frac{_{13}C_3}{_{15C_3}}\right)^3 $

Which comes out to $P(W) = 0.752$

It just seems quite high to me.

So I was messing around with it more and talking to a friend about it. Can anyone tell me if this is correct...

$P(W) = 1 - P(W')$

$ = 1 - \left(\frac{_{13}C_3}{_{15C_3}}\right)^3 $

Which comes out to $P(W) = 0.752$

It just seems quite high to me.

Source Link

So I was messing around with it more and talking to a friend about it. Can anyone tell me if this is correct...

$P(W) = 1 - P(W')$ =

$ = 1 - \left(\frac{_{13}C_3}{_{15C_3}}\right)^3 $

Which comes out to $P(W) = 0.752$

It just seems quite high to me.