Probability of 2 white spheres from 7 spheres?

[Questions999]

Homework Statement

We have 8 white spheres and 5 black spheres in a box.We casually take out of the box a sphere and don't put it there again.Then we take two spheres out of the box.Find the probability that the spheres are both white.

Homework Equations

The Attempt at a Solution

So I put H1 --> the case where we have two white spheres so that means we have 6 white spheres and 5 black spheres
H2- One white sphere and one black sphere : we have 7 white spheres and 4 black spheres
H3 we take have two black spheres so we have 8 white and 3 black spheres

To find P(A) which is what I want I have to find the SUM of P(H)*P(A/H)
I find P(H1)= C (2/8)/C (2/13) and P(A/H1)=C (1/6)/C(1/11)
P(H2)=C (2/5)/ C( 2/13) and P(A/H2)= C (1/7)/ C(1/11)
and P(H3)= C (1/5) * C(1/8)/C (2/13).I find P (A/H3)=C(1/8)/C (1/11)

Now i multiply each P(H) with each P(A/H) and take their sum
BUT something tells me that I am wrong...am I wrong?

 

[Office_Shredder]

What is the event A that you're trying to calculate, and why can't I find anywhere in your post where you take into account that first sphere being removed?

You shouldn't have to calculate the probabilities of H2 and H3 from this problem. Just do it iteratively. Suppose I remove a black ball from the first. What's the probability my second ball is white? What's the probability my third ball is white then?

 

[Ray Vickson]

Homework Statement

We have 8 white spheres and 5 black spheres in a box.We casually take out of the box a sphere and don't put it there again.Then we take two spheres out of the box.Find the probability that the spheres are both white.

Homework Equations

The Attempt at a Solution

So I put H1 --> the case where we have two white spheres so that means we have 6 white spheres and 5 black spheres
H2- One white sphere and one black sphere : we have 7 white spheres and 4 black spheres
H3 we take have two black spheres so we have 8 white and 3 black spheres

To find P(A) which is what I want I have to find the SUM of P(H)*P(A/H)
I find P(H1)= C (2/8)/C (2/13) and P(A/H1)=C (1/6)/C(1/11)
P(H2)=C (2/5)/ C( 2/13) and P(A/H2)= C (1/7)/ C(1/11)
and P(H3)= C (1/5) * C(1/8)/C (2/13).I find P (A/H3)=C(1/8)/C (1/11)

Now i multiply each P(H) with each P(A/H) and take their sum
BUT something tells me that I am wrong...am I wrong?

I cannot make any sense of your approach. Why not take the easy way: look at what happens if the first sphere is white or the first sphere is black (i.e., first = before drawing the next two).

 

[haruspex]

The removal of the first sphere is irrelevant. You just want the probability that two particular spheres (namely, the second and third removed) are both white.

 

[Ray Vickson]

The removal of the first sphere is irrelevant. You just want the probability that two particular spheres (namely, the second and third removed) are both white.

That is one way. Another way is to condition on the color of the first sphere, which then changes the color probabilities of the next two.

More to the point (and maybe very surprising to the OP): the view you suggest leads to the conclusion that the answer is the same as the probability of drawing two balls only and having both white, which is the same as the probability that when we draw three balls the first and third are white, etc. It is even the same as the probability that we draw all 13 balls and find the last two are white!

 

[Questions999]

I looked at this again and changed my solution :
We put H1-> The event when we take the white sphere from the box
We put H2->The event when we take the black sphere from the box
We put A/H1->The event when we take two white spheres,after we have taken a white sphere
and A/H2-->The event when we take two white spheres,after we have taken a black sphere
I find P(H1)=8/13 and P(H2)=5/13
P(A/H1)= C(2/7)/C(2/12) and P(A/H2)=C(2/4)/C(2/12)
We replace this P(A)=P(H1)*P(A/H1) + P(H2)*P(A/H2)..
is this correct?

 

[Ray Vickson]

I looked at this again and changed my solution :
We put H1-> The event when we take the white sphere from the box
We put H2->The event when we take the black sphere from the box
We put A/H1->The event when we take two white spheres,after we have taken a white sphere
and A/H2-->The event when we take two white spheres,after we have taken a black sphere
I find P(H1)=8/13 and P(H2)=5/13
P(A/H1)= C(2/7)/C(2/12) and P(A/H2)=C(2/4)/C(2/12)
We replace this P(A)=P(H1)*P(A/H1) + P(H2)*P(A/H2)..
is this correct?

Is C(a/b) supposed to represent C(b,a) = "b choose a"?

What final numerical value do you get? (I am reluctant to answer your question until you finish your work.)

 

[Questions999]

Ray, C (a/b) means http://s13.postimg.org/ygykmbprb/untitled.png
C is for combinatoric C.

 

[Ray Vickson]

Ray, C (a/b) means http://s13.postimg.org/ygykmbprb/untitled.png
C is for combinatoric C.

That does not help. The usual notations for "b choose a" (b ≥ a) are
$${}_bC_a \text{ or } C_a^b \text{ or } {b \choose a},$$ where the relative position of 'a and 'b' are vital. Your notation seems to have 'a' and 'b' reversed; that is why I asked you the question "do you mean ' b choose a'?" You have still not given me a yes or no answer.

 

[Questions999]

No I mean a choose b.

 

[Ray Vickson]

No I mean a choose b.

OK, so you have C(2/7) = 0, C(2/12) = 0, etc.

 

[Questions999]

No mate, k=2 and n=7 if you know what I mean.

 

[haruspex]

No mate, k=2 and n=7 if you know what I mean.

Then you meant "b choose a", as Ray said. It's short for "from b choose a".