Question 484465
A. drawing a black two out of a standard 52 card deck given it’s not a face card or an Ace.


Since it's "given it's not a face card or an Ace.", we know that the drawn card is a 2, 3, 4, 5, 6, 7, 8, 9, or 10 (and it's either red of black)



There are 9 cards listed from 2 to 10 (shown above) and there are 4 suits. So there are 9*4 = 36 cards that are NOT face cards and that are NOT aces.



So in this part, there are essentially 36 cards to worry about. Note: you can think of it as we're throwing out the face cards and the aces from the deck.



The cards we desire are the two of clubs and the two of spades (both of which are black cards). So there are two cards with the face value of 2 that are black.


So the probability of drawing a black 2 (given the card is NOT a face card and NOT an ace) is {{{2/36=1/18}}}


Note: probability of event = (# of desired outcomes)/(# of total possible out comes)


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B. drawing a six out of a standard 52 card deck given it’s not a face card or an Ace.



Again, we're dealing with 36 cards (we threw out the face cards and aces; see part a)



There are now four cards we desire: a six of hearts, a six of diamonds, a six of spades, and a six of clubs (ie the card with a face value of 6 drawn from each suit)



So the probability is now {{{4/36=1/9}}}


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C. rolling a sum of 7 on two fair six sided dice 


There are 6 ways to roll a 7, and they are: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)


Note: something like (2,5) means that the first die reads "2" and the second die reads "5" (and they add to 2+5=7)



Overall, there are 36 ways to roll 2 dice (since 6*6 = 36). To see this better, make a table where the columns represent die #1 and the rows represent die #2. You'll then see 6*6 = 36 individual boxes (to represent the individual sums of all the possible combos)



So the probability is {{{6/36=1/6}}}


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D. rolling a sum of 3 on two fair six sided dice


There are only 2 ways to roll a 3 and they are: (1,2) and (2,1)



Basically, one die reads "2" and the other reads "1"



From part C, we know that there are 36 different ways two dice can fall.



So the probability is {{{2/36=1/18}}}


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E. rolling a sum of 10 on two fair six sided dice


There are 3 ways to roll a ten and they are: (4,6), (5,5), (6,4)


There are 36 different possible combinations when two dice are rolled.



So the probability is {{{3/36=1/12}}}



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Summary:


We get the following probabilities for each part



A) {{{1/18}}}



B) {{{1/9}}}



C) {{{1/6}}}



D) {{{1/18}}}



E) {{{1/12}}}



So parts A and D have equal probabilities (both equal to {{{1/18}}})



So the answer is    A,D