0
$\begingroup$

I have got some trouble trying to solve this problem that I came up with. I do not even know whether it is solvable or not. So thank you so much if you can help me!

Problem: I have got a number of balls. Some of them are red and some others are black. Moreover some of them are of one size (size 1) and the others of another size (size 2). The only thing I know is that 3/4 of the red are heavy and 2/3 of the black are also heavy.

Question: what is the probability that a ball which is red and of the size 1 is heavy?

Note that size, weight and color are not related by density or volume formulas...

I tried to use the formula $P(A\cap B\cap C)=P(A)P(B|A)P(C|A\cap B)$ but it seems to me there are too many variables.

What do you think?

New contributor
Genus3 is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$
7
  • $\begingroup$ Is "heavy" equivalent to, say, "size $1$"? If not, there is no information about "size" $\endgroup$ Jun 23 at 9:24
  • $\begingroup$ @trueblueanil possibly size $2$ is twice as "big" (radius? volume?) than size $1$ and so heavier $\endgroup$
    – Henry
    Jun 23 at 10:26
  • $\begingroup$ Hey. They are totally uncorrelated. Actually size and weight can deceive... $\endgroup$
    – Genus3
    Jun 23 at 13:29
  • $\begingroup$ If size and weight of red balls are in fact uncorrelated, then the same proportion of red balls size 1 are heavy as the proportion of red balls of size 2, so the answer is $\frac34$. But the question does not say "the size and weight of red balls are uncorrelated". $\endgroup$
    – Henry
    Jun 23 at 13:46
  • $\begingroup$ Uncorrelated in the sense that they are not related by equations involving volume or density... I did not mean to assume proportionality/homogeneity :) $\endgroup$
    – Genus3
    Jun 23 at 15:02

0

Your Answer

Genus3 is a new contributor. Be nice, and check out our Code of Conduct.

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.