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Question Number 141692 by ArielVyny last updated on 22/May/21
Daniel and Bruno are playing with perfect cube  Daniel is the first player if he obtains 1 or 2  he wins the game and the party stopping  or else Bruno plays and if he have {3.4.6} Bruno won and the game stopping  Determine the probability that Daniel winand the probability that Bruno win
$${Daniel}\:{and}\:{Bruno}\:{are}\:{playing}\:{with}\:{perfect}\:{cube} \\ $$$${Daniel}\:{is}\:{the}\:{first}\:{player}\:{if}\:{he}\:{obtains}\:\mathrm{1}\:{or}\:\mathrm{2} \\ $$$${he}\:{wins}\:{the}\:{game}\:{and}\:{the}\:{party}\:{stopping} \\ $$$${or}\:{else}\:{Bruno}\:{plays}\:{and}\:{if}\:{he}\:{have}\:\left\{\mathrm{3}.\mathrm{4}.\mathrm{6}\right\}\:{Bruno}\:{won}\:{and}\:{the}\:{game}\:{stopping} \\ $$$${Determine}\:{the}\:{probability}\:{that}\:{Daniel}\:{winand}\:{the}\:{probability}\:{that}\:{Bruno}\:{win} \\ $$$$ \\ $$
Answered by MJS_new last updated on 22/May/21
in  (1/3) of all cases Daniel wins  in (2/3) of all cases Bruno plays      in (3/6)=(1/2) of these cases Bruno wins      the rest end in a draw  ⇒  Daniel wins (1/3)  Bruno wins (2/3)×(1/2)=(1/3)  draw (1/3)
$$\mathrm{in}\:\:\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{of}\:\mathrm{all}\:\mathrm{cases}\:\mathrm{Daniel}\:\mathrm{wins} \\ $$$$\mathrm{in}\:\frac{\mathrm{2}}{\mathrm{3}}\:\mathrm{of}\:\mathrm{all}\:\mathrm{cases}\:\mathrm{Bruno}\:\mathrm{plays} \\ $$$$\:\:\:\:\mathrm{in}\:\frac{\mathrm{3}}{\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{of}\:\mathrm{these}\:\mathrm{cases}\:\mathrm{Bruno}\:\mathrm{wins} \\ $$$$\:\:\:\:\mathrm{the}\:\mathrm{rest}\:\mathrm{end}\:\mathrm{in}\:\mathrm{a}\:\mathrm{draw} \\ $$$$\Rightarrow \\ $$$$\mathrm{Daniel}\:\mathrm{wins}\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{Bruno}\:\mathrm{wins}\:\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{draw}\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$

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