Question #225993

Two 4-sided dice are tossed and the sum of the down sides noted. A 4-sided die doesn’t have an “up” face like a 6-sided die has, so the down side is used. Determine the expected value of the sum. Provide evidence of your work by typing out your full solution. 


Expert's answer

Below is probability distribution for sum of  down sides :

P(X=2)=P(both sides show 1)=116(one outcome (1,1)out of44=16 outcomes)P(X=3)=216=18(two outcomes(1,2),(2,1))P(X=4)=316(three outcomes(1.3),(2,2),(3,1))P(X=5)=416=14(four outcomes(1.4),(2,3),(3.2),(4,1))P(X=6)=316(three outcomes(2,4),(3,3),(4,2))P(X=7)=216=18(two outcomes(3,4),(4,3))P(X=8)=116(one such outcome(4,4))P(X=2) =P(both \space sides \space show \space 1) =\frac{1}{16} \\ (one \space outcome \space (1,1) out \space of 4*4 =16 \space outcomes)\\ P(X=3) =\frac{2}{16} =\frac{1}{8} (two \space outcomes (1,2),(2,1))\\ P(X=4) =\frac{3}{16} (three \space outcomes (1.3),(2,2),(3,1))\\ P(X=5)=\frac{4}{16} =\frac{1}{4} (four \space outcomes (1.4),(2,3),(3.2),(4,1))\\ P(X=6) =\frac{3}{16} (three \space outcomes (2,4),(3,3),(4,2))\\ P(X=7)=\frac{2}{16} =\frac{1}{8} (two \space outcomes (3,4),(4,3))\\ P(X=8)=\frac{1}{16} (one \space such \space outcome (4,4))\\


The expected value

E(s)=i=1Nsip(si)=2(116)+3(216)+4(316)+5(416)+6(316)+7(216)+8(116)=(8016)=5E(s)= \sum_{i=1}^{N}s_ip(s_i)\\ =2(\frac{1}{16})+3(\frac{2}{16})+4(\frac{3}{16})+5(\frac{4}{16})+6(\frac{3}{16})+7(\frac{2}{16})+8(\frac{1}{16})\\ = (\frac{80}{16})\\ =5


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