bag , gold and probability

Even numbers can be grouped into three groups : $6n, 6n+2, 6n+4$ .

Since we are dealing with real entities like gold coins, $n$ is a positive integer.

So there is only one element out of the three which is a multiple of $6$ , the required probability is $\boxed {\dfrac{1}{3}}$ .

We have to find the probability of having a multiple of 6 number of coins given that Jack have even number of coins. Let the probability of a number to be even be $P(E)$ and probability of a number to be multiple of 6 be $P(S)$ . Then

$P(E) = \large\frac{1}{2}$ and $P(S) = \large\frac{1}{6}$

We have to find $P(S/E)$

$P(S/E) = \large\frac{P(S\cap E)}{P(E)}$ $= \large \frac{P(S)}{P(E)}$ $= \large\frac{\frac{1}{6}}{\frac{1}{2}} = \frac{1}{3}$ $\hspace{30pt} [\because S\cap E = S]$