There are 7 persons. 4 with a master degree and 3 without a master degree. If you ran

[FONT=&quot]Hi, can someone help me how to do this:[/FONT]
[FONT=&quot]There are 7 persons. 4 with a master degree and 3 without a master degree. If you randomly select 3 persons, what is the probability distribution of the number of people selected in the experiment that have a master degree.[/FONT]

neto333 said:

Hi, can someone help me how to do this:
There are 7 persons. 4 with a master degree and 3 without a master degree. If you randomly select 3 persons, what is the probability distribution of the number of people selected in the experiment that have a master degree.

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What are your thoughts? What have you tried? Where are you stuck?

Please be complete. Thank you!

stapel said:

What are your thoughts? What have you tried? Where are you stuck?

Please be complete. Thank you!

Click to expand...

HI, i tried with combinatorics this is what i got:
p(x=0)=(4!/(0!*4!))/(7!/(3!*4!))=1/35
p(x=1)=(4!/(1!*3!))/(7!/(3!*4!))=4/35
p(x=2)=6/35
p(x=3)=4/35
but then when i sum all the probabilities, the sum is not one.
So then i tried another way
p(x=0)=3/7*2/6*1/5=1/35
p(x=1)=4/7*3/6*2/5=4/35
But is the same... so basically im not sure what to do.

stapel said:

What are your thoughts? What have you tried? Where are you stuck?

Please be complete. Thank you!

Click to expand...

HI, i tried with combinatorics this is what i got:
p(x=0)=(4!/(0!*4!))/(7!/(3!*4!))=1/35p(x=1)=(4!/(1!*3!))/(7!/(3!*4!))=4/35p(x=2)=6/35p(x=3)=4/35but then when i sum all the probabilities, the sum is not one.So then i tried another way: p(x=0)=3/7*2/6*1/5=1/35p(x=1)=4/7*3/6*2/5=4/35 But is the same... so basically im not sure what to do.

I'll assume that you're using "x" to stand for "the number of people with a master's degree".

neto333 said:

i tried with combinatorics this is what i got:
p(x=0)=(4!/(0!*4!))/(7!/(3!*4!))=1/35
p(x=1)=(4!/(1!*3!))/(7!/(3!*4!))=4/35

Click to expand...

Clearly, yes, there is only one way to pick zero people with a master's degree: pick all three of the people (and only those people) without said degree.

However, are there really only four ways to pick one person with a master's degree? If we label the people as:

. . . . .without: a, b, and c

. . . . .with: d, e, f, and g

...then we can pick as follows:

. . . . .a, b, and d
. . . . .a, b, and e
. . . . .a, b, and f
. . . . .a, b, and g
. . . . .a, c, and d
. . . . .a, c, and e
. . . . .a, c, and f
. . . . .a, c, and g
. . . . .b, c, and d
. . . . .b, c, and e
. . . . .b, c, and f
. . . . .b, c, and g

That's twelve ways, not just four! So check your reasoning and your formulas.

neto333 said:

Hi, can someone help me how to do this:
There are 7 persons. 4 with a master degree and 3 without a master degree. If you randomly select 3 persons, what is the probability distribution of the number of people selected in the experiment that have a master degree.

Click to expand...

Assume that you're using "x" to stand for "the number of people with a master's degree".
Then x can have four values. What are those values?
\(\displaystyle P(x)=\dfrac{\mathcal{C^4_x}\cdot\mathcal{C^3_{3-x}}}{\mathcal{C^7_3}}\)