Question

An urn contains 4 red balls and 6 blue balls. A second urn contains 16 red balls and an unknown amount of blue balls. If a single ball is drawn from each urn, the probability that both balls are the same color is 0.44. How many blue balls are in the second urn?

Answer 1

harder than i first thought, lets see if we can do it
put the number of blue balls as \(x\)
then the probability the first ball and the second ball is blue is
\[.6\times \frac{x}{16+x}\] and the probability that the first and second ball chosen is red is
\[.4\times \frac{16-x}{16+x}\]

the total is \(0.44\) so we can set
\[\frac{.6x}{16+x}+\frac{.4(16-x)}{16+x}=0.44\] and try to solve for \(x\)

Answer 2

oh no that was wrong sorry

Answer 3

the probability that both chosen are red is
\[.4\times \frac{16}{16+x}\]

Answer 4

so set
\[\frac{.6x}{16+x}+\frac{.4\times 16}{16+x}=0.44\]

Answer 5

\[\frac{.6x+6.4}{16+x}=0.44\]
\[.6x+6.4=0.44(16+x)\] etc