20 different beads, 4 out of them are yellow, choose 7, Calc P(exactly 2 of the 7 are yellow)

[Alon189]

We have 20 different beads. 4 of them are yellow. We choose and take one after the other 7 beads, what is the probability that exactly 2 of them are yellow?

What i did:

We want exactly 2 yellow so \[ \binom{20}{2} \] and the rest must not be yellow (cuz its exactly 2) therefore we are left with 16 from them we choose 5, namely: \[\binom{16}{5}\] the total amount of beads we took are \[\binom{20}{7}\].

Therefore i thought that the probability we are looking for is:
\[P = \frac{\binom{20}{2}\cdot \binom{16}{5}}{\binom{20}{7}}\].

Surely its not the answer (I get P>1), but why do i wrong?

Thanks.

 

[joshuaa]

[MATH]\frac{{4 \choose 2} {16 \choose 5} }{{20 \choose 7} }[/MATH]

 

[Steven G]

Note heavily that in joshuaa's answer in the numerator the top numbers add up to 4+16 =20 and the lower numbers add up to 2+5=7 which are exactly the numbers in the denominator! If that doesn't happen then go and find your error.

You are NOT choosing 2 yellows from 20, rather you are choosing 2 yellows from 4. After all there are NOT 20 yellows to choose from!

 

[joshuaa]

Note heavily that in joshuaa's answer in the numerator the top numbers add up to 4+16 =20 and the lower numbers add up to 2+5=7 which are exactly the numbers in the denominator! If that doesn't happen then go and find your error.

You are NOT choosing 2 yellows from 20, rather you are choosing 2 yellows from 4. After all there are NOT 20 yellows to choose from!

lol

then you Understand nothing in probability

Jomo, you forced me to laugh. if you don't understand probability, don't make people make fun of you

 

[Alon189]

lol

then you Understand nothing in probability

Jomo, you forced me to laugh. if you don't understand probability, don't make people make fun of you

This forum has no rules against unpleasant replys and bullying?

 

[Steven G]

lol

then you Understand nothing in probability

Jomo, you forced me to laugh. if you don't understand probability, don't make people make fun of you

When using hypergeometric what I wrote is most certainly true. Are you disagreeing with what I said? It is a good check to see if you must be wrong. What are you laughing at?

 

[joshuaa]

forgive me Jomo. i misunderstood your comment. i thought you are telling Alon that my answer is wrong

it's whether i don't understand English or i did not sleep well

 

[Steven G]

forgive me Jomo. i misunderstood your comment. i thought you are telling Alon that my answer is wrong

it's whether i don't understand English or i did not sleep well

You're forgiven