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We have 8 white spheres and 5 black spheres in a box.We casually take out of the box a sphere and dont put it there again.Then we take two spheres out of the box.Find the probability that the spheres are both white.

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 im wrong...am I wrong?

EDIT :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?

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Are you first taking out a single sphere and then drawing two more (three spheres in total), or do you only draw two spheres in total? –  Austin Mohr Jun 4 '13 at 8:10
    
I am first taking a single sphere,and then taking two.. –  hgdhg Jun 4 '13 at 10:52
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1 Answer 1

I take your question to mean what is the probability of BWW or WWW.

P(WWW) = 8/13 times 7/12 times 6/11 = 336/1716

P(BWW) = 5/13 times 8/12 times 7/11 = 280/1716

So Answer = P(WWW) + P(BWW) = (336 + 280)/1716 = 616/1716 = 0.3590

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