2006年AMC10竞赛B第21题
对于一对特殊的骰子，在每个骰子上掷出$1$、$2$、$3$、$4$、$5$和$6$的概率的比例是$1:2:3:4:5:6$.两颗骰子共掷出$7$的概率是多少？
课内体系->知识点->统计与概率->概率->事件与概率->互斥事件->互斥事件的概率加法公式 美国AMC10/12->Knowledge Point->Combination->Probability Calculation->Classical Models of Probabilities
Let $$x$$ be the \uline{probability} of rolling a $$1$$. The probabilities of rolling a $$2$$, $$3$$, $$4$$, $$5$$, and $$6$$ are $$2x$$, $$3x$$, $$4x$$, $$5x$$, and $$6x$$, respectively.  The sum of the probabilties of rolling each number must equal $$1$$, so  $$x+2x+3x+4x+5x+6x=1$$,  $$21x=1$$,  $$x= \dfrac{1}{21}$$,  So the probabilities of rolling a $$1$$, $$2$$, $$3$$, $$4$$, $$5$$, and $$6$$ are respectively $$\dfrac{1}{21}$$,~ $$\dfrac{2}{21}$$, $$\dfrac{3}{21}$$, $$\dfrac{4}{21}$$, $$\dfrac{5}{21}$$, and, $$\dfrac{6}{21}$$. The possible combinations of two rolls that total $$7$$ are: $$(1,6)$$; $$(2,5)$$; $$(3,4)$$; $$(4,3)$$; $$(5,2)$$; $$(6,1)$$. The probability of rolling a total of $$7$$ on the two dice is equal to the sum of the probabilities of rolling each combination. $$P= \dfrac{1}{21}\cdot \dfrac{6}{21}+ \dfrac{2}{21}\cdot \dfrac{5}{21}+ \dfrac{3}{21}\cdot \dfrac{4}{21}+ \dfrac{4}{21}\cdot \dfrac{3}{21}+ \dfrac{5}{21}\cdot \dfrac{2}{21}+ \dfrac{6}{21}\cdot \dfrac{1}{21}= \dfrac{8}{63}\Rightarrow C$$ 