Question #233141

Roll both a 10-sided fair die with an Ace side and a 12-sided fair die with an Ace side.


(a) What is the chance that one, and only one, of the dice will land Ace?  (Enter exact answer.)


(b) Given that one, and only one, of the dice landed Ace, what is the probability it was the 10-sided die?  (Enter exact answer.)


Expert's answer

(a)


P(only one Ace)=110(1112)+910(112)P(only\ one\ Ace)=\dfrac{1}{10}(\dfrac{11}{12})+\dfrac{9}{10}(\dfrac{1}{12})

=20120=16=\dfrac{20}{120}=\dfrac{1}{6}

(b)


P(10sidedonly one Ace)P(10-sided|only\ one\ Ace)

=P(only one Ace10sided)P(only one Ace)=\dfrac{P(only\ one\ Ace|10-sided)}{P(only\ one\ Ace)}

=110(1112)16=1120=\dfrac{\dfrac{1}{10}(\dfrac{11}{12})}{\dfrac{1}{6}}=\dfrac{11}{20}



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