Problem of coins There is a wallet that contains: 5 quarters, 3 dimes, 8 nickels, and 4 pennies. A coin is drawn from the purse and replaced 240 times. How many times can you predict that a nickel or a penny will be drawn?
Could you give me an idea?
 A: A nickel or penny will be drawn with probability $\frac{8+4}{5+3+8+4}=\frac{12}{20}=\frac{3}{5}$. These can be interpreted as independent Bernoulli trials with $p=.6$. The expected number of times a nickel or penny will be drawn is $240 \times .6=144$.
A: We divide the coins in three variables: quarters drawn $X_1$, dimes drawn $X_2$, and pennies+nickels drawn $X_3$. The distribution is multinomial with probabilities:
$$p_1=\frac{5}{20} \ \ \ p_2=\frac{3}{20} \ \ \ p_3=\frac{12}{20}$$
and the pmf for $n$ trials and $x_1,x_2,x_3$ successes such that $\sum^3_{k=1}x_k=n$ is
$$P(X_1=x_1,X_2=x_2,X_3=x_3)=\frac{n!}{x_1!x_2!x_3!}p_1^{x_1}p_2^{x_2}p_3^{x_3}=\frac{n!}{(n-x_2-x_3)!x_2!x_3!}p_1^{n-x_2-x_3}p_2^{x_2}p_3^{x_3}$$
So we have to marginalize over $X_2$ obtaining
$$P(X_3=x_3)=\frac{n!}{x_3!(n-x_3)!}p_3^{x_3}(1-p_3)^{n-x_3}, \ \ \ x_3\in \mathbb{N}_0$$
which is similar to a binomial pmf. The expected times a nickel or a penny will be drawn are
$$\mathbb{E}[X_3]=np_3$$
that is $240\cdot \frac{12}{20}=144$
