13. (1980 American High School Mathematics Exam Question) A box contains 2 one-cent coins, 4 five-cent coins, and 6 ten-cent coins. Six coins are drawn sequentially from the box without replacement, with each coin having an equal probability of being selected. The probability that the value of the drawn coins is at least 50 cents is ( ).
A. $\frac{37}{924}$
B. $\frac{91}{924}$
C. $\frac{127}{924}$
D. $\frac{132}{924}$
E. None of these
13. C. Reason: The number of ways to choose 6 coins from 12 is $C_{12}^{6}=924$. The following situations will result in an event of "at least 50 cents":
(1) Taking out 6 dimes;
(2) Taking out 5 dimes and any other coin;
(3) Taking out 4 dimes and 2 nickels;

The number of ways for (1), (2), and (3) to occur are $C_{6}^{5}, C_{6}^{6} \cdot C_{6}^{1}$, and $C_{6}^{4} \cdot C_{4}^{2}$, respectively.
Therefore, the required probability is $\frac{C_{6}^{6}+C_{6}^{4} \cdot C_{4}^{2}+C_{6}^{6} \cdot C_{6}^{1}}{924}=\frac{127}{924}$.