Question

A bin has 5 white balls and k black balls in it, where k is an unknown positive integer. A ball is drawn at random from the bin. If a white ball is drawn, the player wins 1 dollar, but if a black ball is drawn, the player loses 1 dollar. If the expected loss for playing the game is 50 cents, then what is k?

Answer 1

okay, so if on average we expect to lose \(1/2\) of a dollar, that means we must have more black balls than white balls (so that it's more than likely to lose money). in fact, with a little bit of thinking, you should realize that the \(5\) balls must be a quarter of the total amount of balls, meaning that \(k\) ought to be \(3\times5=15\)...

Answer 2

that's the quick intuitive way of solving the problem. alternatively, let's write out our expected loss as a negative expected gain i.e. \(-1/2\) and consider:$$E(X)=p_1x_1+p_2x_2+\dots+p_3x_n$$for a discrete random variable \(X\in\{x_1,x_2,\dots x_n\}\) with corresponding probability mass distribution given by \(p_1,p_2,\dots,p_n\). Here, we have \(x_1=1,x_2=-1\) for our expected values of gain (either we gain a dollar i.e. \(+1\) or we lose one i.e. \(-1\)). the probabilities given it is a simple random sample should just be given by the ratio of the numbers of white (\(5\))/black (\(k\)) balls to the total number of balls, \(5+k\) so \(p_1=5/(5+k),p_2=k/(5+k)\) so our expected value gives th following equation:$$E(X)=\frac5{5+k}-\frac{k}{5+k}\\\frac12=\frac{5-k}{5+k}\\-\frac12=\frac{5+k-2k}{5+k}\\-\frac12=1-\frac{2k}{5+k}\\\frac{2k}{5+k}=1+\frac12=\frac32\\4k=3(5+k)=15+3k\\k=15$$

Answer 3

Thanks, I understand it now.