Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 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?

share|cite|improve this question
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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.