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Question Number 101121 by byaw last updated on 30/Jun/20
Two dice are tossed simultaneously.  Find the probability of any showing  a number greater than five.
$$\mathrm{Two}\:\mathrm{dice}\:\mathrm{are}\:\mathrm{tossed}\:\mathrm{simultaneously}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{of}\:\mathrm{any}\:\mathrm{showing} \\ $$$$\mathrm{a}\:\mathrm{number}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{five}. \\ $$
Answered by MJS last updated on 30/Jun/20
⇒ dice A or dice B = 6  =  not (dice A and dice B both ≠6) =  =1−((5/6))^2 =((11)/(36))  11 21 31 41 51 61  12 22 32 42 52 62  13 23 33 43 53 63  14 24 34 44 54 64  15 25 35 45 55 65  16 26 36 46 56 66
$$\Rightarrow\:\mathrm{dice}\:\mathrm{A}\:{or}\:\mathrm{dice}\:\mathrm{B}\:=\:\mathrm{6} \\ $$$$= \\ $$$$\mathrm{not}\:\left(\mathrm{dice}\:\mathrm{A}\:{and}\:\mathrm{dice}\:\mathrm{B}\:\mathrm{both}\:\neq\mathrm{6}\right)\:= \\ $$$$=\mathrm{1}−\left(\frac{\mathrm{5}}{\mathrm{6}}\right)^{\mathrm{2}} =\frac{\mathrm{11}}{\mathrm{36}} \\ $$$$\mathrm{11}\:\mathrm{21}\:\mathrm{31}\:\mathrm{41}\:\mathrm{51}\:\mathrm{61} \\ $$$$\mathrm{12}\:\mathrm{22}\:\mathrm{32}\:\mathrm{42}\:\mathrm{52}\:\mathrm{62} \\ $$$$\mathrm{13}\:\mathrm{23}\:\mathrm{33}\:\mathrm{43}\:\mathrm{53}\:\mathrm{63} \\ $$$$\mathrm{14}\:\mathrm{24}\:\mathrm{34}\:\mathrm{44}\:\mathrm{54}\:\mathrm{64} \\ $$$$\mathrm{15}\:\mathrm{25}\:\mathrm{35}\:\mathrm{45}\:\mathrm{55}\:\mathrm{65} \\ $$$$\mathrm{16}\:\mathrm{26}\:\mathrm{36}\:\mathrm{46}\:\mathrm{56}\:\mathrm{66} \\ $$

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