AMC10 2006 B

AMC10 2006 B · Q21

AMC10 2006 B · Q21. It mainly tests Fractions, Probability (basic).

For a particular peculiar pair of dice, the probabilities of rolling 1, 2, 3, 4, 5, and 6 on each die are in the ratio 1 : 2 : 3 : 4 : 5 : 6. What is the probability of rolling a total of 7 on the two dice?

对于一对特殊的骰子，每颗骰子上掷出1、2、3、4、5和6的概率之比为1 : 2 : 3 : 4 : 5 : 6。掷两个骰子总和为7的概率是多少？

(A) $\frac{4}{63}$ $\frac{4}{63}$

(B) $\frac{1}{8}$ $\frac{1}{8}$

(D) $\frac{1}{6}$ $\frac{1}{6}$

(E) $\frac{2}{7}$ $\frac{2}{7}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

(C) On each die the probability of rolling $k$, for $1 \le k \le 6$, is \[ \frac{k}{1+2+3+4+5+6}=\frac{k}{21}. \] There are six ways of rolling a total of 7 on the two dice, represented by the ordered pairs $(1,6)$, $(2,5)$, $(3,4)$, $(4,3)$, $(5,2)$, and $(6,1)$. Thus the probability of rolling a total of 7 is \[ \frac{1\cdot 6+2\cdot 5+3\cdot 4+4\cdot 3+5\cdot 2+6\cdot 1}{21^2} =\frac{56}{21^2} =\frac{8}{63}. \]

（C）对每个骰子而言，掷出$k$（其中$1 \le k \le 6$）的概率为 \[ \frac{k}{1+2+3+4+5+6}=\frac{k}{21}. \] 两个骰子点数和为7的情况有6种，对应有序对$(1,6)$、$(2,5)$、$(3,4)$、$(4,3)$、$(5,2)$和$(6,1)$。因此，点数和为7的概率为 \[ \frac{1\cdot 6+2\cdot 5+3\cdot 4+4\cdot 3+5\cdot 2+6\cdot 1}{21^2} =\frac{56}{21^2} =\frac{8}{63}. \]

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