Question 462475: PLEASE HELP ME, I posted this problem once, and no one responded, can you help me?
which pair has equally likley outcomes? list the letters of the two choices below which have equal probabilities of success.
a)drawing a red seven out of a standard 52 card deck given its not a face card or an ace.
b)drawing a five out of a standard 52 card deck given its not a face card or an ace.
c)rolling a sum of 4 on two fair six sided dice.
d)rolling a sum of 5 on two fair six sided dice.
e)rolling a sum of 7 on two fair six sided dice.
Answer by math-vortex(648)   (Show Source):
You can put this solution on YOUR website! --------------------------
a)drawing a red seven out of a standard 52 card deck given its not a face card or an ace.
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P(red 7|not face or ace) = P(not face or ace AND red 7) / P(not face or ace)
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P(red seven AND not face or ace) = 2/52 because there are 2 red seven cards (red-diamond and red-heart) in the deck of 52 cards.
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P(not face or ace) = 36/52 because there are 9*4=36 non-face or non-ace cards (2 through 10 of every suit) in a deck of 52 cards.
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P(red 7|not face or ace) = (2/52) / (36/52) = 2/36
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b)drawing a five out of a standard 52 card deck given its not a face card or an ace.
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P(five|not face or ace) = P(not face or ace AND red 7) / P(not face or ace)
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P(five AND not face or ace) = 4/52 because there are 4 fives (five-diamond, -heart, -club, and -spade) in the deck of 52 cards.
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P(not face or ace) = 36/52 because there are 9*4=36 non-face or non-ace cards (2 through 10 of every suit) in a deck of 52 cards.
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P(five|not face or ace) = (4/52) / (36/52) = 4/36
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more coming soon!!!
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c)rolling a sum of 4 on two fair six sided dice.
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A good way to solve die rolling problems is to build a grid of possible dice sums.
+ 1 2 3 4 5 6
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1| 2 3 4 5 6 7
2| 3 4 5 6 7 8
3| 4 5 6 7 8 9
4| 5 6 7 8 9 10
5| 6 7 8 9 10 11
6| 7 8 9 10 11 12
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(hopefully, this lines up nicely in your browser (o:
Now we can see count easily.
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Events:
E = rolling a sum of 4
S = rolling any sum
There are 3 ways to roll a sum of 4 [(1+3, 3+1, 2+2)] and 36 ways to roll ANY sum. So the probability of our event is
P(E) = n(E) / n(S) = 3/36
Problems (d) and (e) use identical reasoning to (c). They will be good practice for you. Finish the problem by finding the two events with equal probabilities.
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hope this helps! Feel free to contact me at math.in.the.vortex@gmail.com.
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Ms.Figgy
math.in.the.vortex
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