You are a trader considering a stock whose next price movement will be up or down with equal probability. Fortunately, you have access to a binary signal that is predictive of the price change (and observable ahead of time). The signal is 60% accurate: conditional on the stock going up, there's a 0.6 probability that the signal says up (and 0.4 probability that it says down); the reverse is true if the stock goes down.
You have access to many such signals, all with the same accuracy. Assume they are conditionally independent given the price movement (e.g. conditional on the stock going up, there's a 0.6^2 probability that signals one and two both point up). Suppose you observe N+1 signals pointing up and N pointing down (out of a total of 2N+1 signals, i.e. the total number must be odd). What's the probability that the stock will go up?
Let $\ M\ $ be the random variable indicating the movement in the stock price, $\ S_i, i = 1, 2, \dots, 2\, N+1\ $, the signals, and $\ U, D\ $ the sets of indices $\ i\ $ for which the signal $\ S_i\ $ has the value $\ u\ $ (for "up") or $\ d\ $ (for "down"), respectively. We are told that $\ \vert U\vert = N+1\ $ and $\ \vert D\vert = N\ $.
\begin{eqnarray} \ & &\mathrm{P}\left(\,M = u\,\vert\, S_i = u\ \mathrm{for}\ i\in U\ \&\ S_i = d\ \mathrm{for}\ i\in D\, \right) \\ &=& \frac{\mathrm{P}\left(\,M = u\,\&\, S_i = u\ \mathrm{for}\ i\in U\ \&\ S_i = d\ \mathrm{for}\ i\in D\, \right)}{\mathrm{P}\left(\, S_i = u\ \mathrm{for}\ i\in U\ \&\ S_i = d\ \mathrm{for}\ i\in D\, \right)}\\ &=& \frac{\mathrm{P}\left(\, S_i = u\ \mathrm{for}\ i\in U\ \&\ S_i = d\ \mathrm{for}\ i\in D\,\vert\,M=u\, \right)\,\mathrm{P}\left(\,M=u\,\right)}{\mathrm{P}\left(\, S_i = u\ \mathrm{for}\ i\in U\ \&\ S_i = d\ \mathrm{for}\ i\in D\, \right)}\\ &=& \frac{0.6^{N+1}\,0.4^N\,0.5}{\mathrm{P}\left(\, S_i = u\ \mathrm{for}\ i\in U\ \&\ S_i = d\ \mathrm{for}\ i\in D\, \right)}\ . \end{eqnarray} For the denominator, we have \begin{eqnarray} & &\mathrm{P}\left(\, S_i = u\ \mathrm{for}\ i\in U\ \&\ S_i = d\ \mathrm{for}\ i\in D\, \right)\\ &=& \mathrm{P}\left(\, S_i = u\ \mathrm{for}\ i\in U\ \&\ S_i = d\ \mathrm{for}\ i\in D\,\vert\,M=u \right)\mathrm{P}\left(\,M=u\,\right)\\ & & +\, \mathrm{P}\left(\, S_i = u\ \mathrm{for}\ i\in U\ \&\ S_i = d\ \mathrm{for}\ i\in D\,\vert\,M=d \right)\mathrm{P}\left(\,M=d\,\right)\\ &=& 0.6^{N+1}\,0.4^N\,0.5 + 0.6^N\,0.4^{N+1}\,0.5\\ &=& 0.6^N\,0.4^N\,0.5\ . \end{eqnarray} So finally, \begin{eqnarray} \ & &\mathrm{P}\left(\,M = u\,\vert\, S_i = u\ \mathrm{for}\ i\in U\ \&\ S_i = d\ \mathrm{for}\ i\in D\, \right)\\ &=& \frac{0.6^{N+1}\,0.4^N\,0.5}{0.6^N\,0.4^N\,0.5} = 0.6 \end{eqnarray}