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andijo76
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Which pair has equally likely outcomes? List the letters of the two choices below which have equal probabilities of success, separated by a comma. A standard deck of cards has 12 face cards and four Aces (Aces are not face cards).
A. rolling a sum of 4 on two fair six sided dice
B. drawing a black seven out of a standard 52 card deck given it’s not a face card or an Ace.
C. rolling a sum of 11 on two fair six sided dice
D. rolling a sum of 7 on two fair six sided dice
E. drawing a three out of a standard 52 card deck given it’s not a face card or an Ace.
 2 years ago
 2 years ago
andijo76 Group Title
Which pair has equally likely outcomes? List the letters of the two choices below which have equal probabilities of success, separated by a comma. A standard deck of cards has 12 face cards and four Aces (Aces are not face cards). A. rolling a sum of 4 on two fair six sided dice B. drawing a black seven out of a standard 52 card deck given it’s not a face card or an Ace. C. rolling a sum of 11 on two fair six sided dice D. rolling a sum of 7 on two fair six sided dice E. drawing a three out of a standard 52 card deck given it’s not a face card or an Ace.
 2 years ago
 2 years ago

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satellite73 Group TitleBest ResponseYou've already chosen the best response.1
wow this is maybe the 4th time i have seen this exact problem. where does it come from?
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
we can do the dice first \[P(A)=\frac{3}{36}=\frac{1}{12}\] since you can roll (3,1), (2,2), (1,3)
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
\[P(C)=\frac{2}{36}=\frac{1}{18}\] possible rolls are (5,6), (6,5)
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
\[P(D)=\frac{6}{36}=\frac{1}{6}\] and that takes care of the dice
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
deck has 12 face cards and 4 aces so you know it is not one of those 16 cards and must be one of the remaining 36 cards. there are 2 black sevens, so probability of picking a 7 if you know it is not one of the above 16 cards is \[\frac{2}{36}=\frac{1}{18}\]
 2 years ago
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