Question 484465: I just confused my school math teacher! How lovely! Anyways HOPE you guys can seriously help me out more then these people at my college! I dont get why I pay thousands of dollars and not get adequate help!
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. drawing a black two out of a standard 52 card deck given it’s not a face card or an Ace.
B. drawing a six out of a standard 52 card deck given it’s not a face card or an Ace.
C. rolling a sum of 7 on two fair six sided dice
D. rolling a sum of 3 on two fair six sided dice
E. rolling a sum of 10 on two fair six sided dice
Found 2 solutions by chessace, jim_thompson5910:
Answer by chessace(471)   (Show Source):
You can put this solution on YOUR website! The wording is a little weird.
The deck of cards is essentially reduced to 36.
All the probabilities are /36:
A=2 B=4 C=6 D=2 E=3
So A=D
Answer by jim_thompson5910(35256)  (Show Source):
You can put this solution on YOUR website! 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 
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 
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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 
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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
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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 
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Summary:
We get the following probabilities for each part
A) 
B) 
C) 
D)
E) 
So parts A and D have equal probabilities (both equal to )
So the answer is A,D
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