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Question-177008




Question Number 177008 by Ar Brandon last updated on 29/Sep/22
Answered by Rasheed.Sindhi last updated on 29/Sep/22
Let t is total of card numbers chose by players.  Efe′s card number=3         6≤t≤20  If t∈P then possible values for t:       t=7,11,13,17,19  total of other two′s:  t−3=4,8,10,14,16  Pairs of other two:   determinant (((t−3),(possible pairs)),(4,−),(8,((1,7),(2,6),(6,2),(7,1))),((10),((1,9),(2,8),(4,6),(6,4),(8,2),(9,1))),((14),((5,9),(6,8),(8,6),(9,5))),((16),((7,9),(9,7))))  Total possible pairs for other two:  8×7=56  Number of pairs when Efe wins=16  probability to win Efe:         =((16)/(56))=(2/7)
$${Let}\:{t}\:{is}\:{total}\:{of}\:{card}\:{numbers}\:{chose}\:{by}\:{players}. \\ $$$${Efe}'{s}\:{card}\:{number}=\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\mathrm{6}\leqslant{t}\leqslant\mathrm{20} \\ $$$${If}\:{t}\in\mathbb{P}\:{then}\:{possible}\:{values}\:{for}\:{t}: \\ $$$$\:\:\:\:\:{t}=\mathrm{7},\mathrm{11},\mathrm{13},\mathrm{17},\mathrm{19} \\ $$$${total}\:{of}\:{other}\:{two}'{s}: \\ $$$${t}−\mathrm{3}=\mathrm{4},\mathrm{8},\mathrm{10},\mathrm{14},\mathrm{16} \\ $$$${Pairs}\:{of}\:{other}\:{two}: \\ $$$$\begin{array}{|c|c|c|c|c|c|}{{t}−\mathrm{3}}&\hline{{possible}\:{pairs}}\\{\mathrm{4}}&\hline{−}\\{\mathrm{8}}&\hline{\left(\mathrm{1},\mathrm{7}\right),\left(\mathrm{2},\mathrm{6}\right),\left(\mathrm{6},\mathrm{2}\right),\left(\mathrm{7},\mathrm{1}\right)}\\{\mathrm{10}}&\hline{\left(\mathrm{1},\mathrm{9}\right),\left(\mathrm{2},\mathrm{8}\right),\left(\mathrm{4},\mathrm{6}\right),\left(\mathrm{6},\mathrm{4}\right),\left(\mathrm{8},\mathrm{2}\right),\left(\mathrm{9},\mathrm{1}\right)}\\{\mathrm{14}}&\hline{\left(\mathrm{5},\mathrm{9}\right),\left(\mathrm{6},\mathrm{8}\right),\left(\mathrm{8},\mathrm{6}\right),\left(\mathrm{9},\mathrm{5}\right)}\\{\mathrm{16}}&\hline{\left(\mathrm{7},\mathrm{9}\right),\left(\mathrm{9},\mathrm{7}\right)}\\\hline\end{array} \\ $$$$\mathcal{T}{otal}\:{possible}\:{pairs}\:{for}\:{other}\:{two}: \\ $$$$\mathrm{8}×\mathrm{7}=\mathrm{56} \\ $$$$\mathcal{N}{umber}\:{of}\:{pairs}\:{when}\:{Efe}\:{wins}=\mathrm{16} \\ $$$${probability}\:{to}\:{win}\:{Efe}: \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{16}}{\mathrm{56}}=\frac{\mathrm{2}}{\mathrm{7}} \\ $$
Commented by Ar Brandon last updated on 29/Sep/22
Thank you, Sir !
$$\mathrm{Thank}\:\mathrm{you},\:\mathrm{Sir}\:! \\ $$
Commented by Tawa11 last updated on 02/Oct/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

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