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. 


1
Expert's answer
2021-08-17T09:57:54-0400

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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