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Difference between revisions of "Suppose we flip four coins simultaneously: a penny, a nickel, a dime, and a quarter. What is the probability that the penny and dime both come up the same?"

(Created page with "There are <math>2^4=16</math> possible outcomes, since each of the 4 coins can land 2 different ways (heads or tails). There are 2 possibilities for the penny and the dime: ei...")
 
(added deletion request)
 
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There are <math>2^4=16</math> possible outcomes, since each of the 4 coins can land 2 different ways (heads or tails). There are 2 possibilities for the penny and the dime: either they're both heads or they're both tails. There are also 2 possibilities for the nickel and 2 possibilities for the quarter. So there are <math>2 \times 2 \times 2 = 8</math> successful outcomes, and the probability of success is <math>\dfrac{8}{16} = \boxed{\dfrac{1}{2}}</math>.
 
There are <math>2^4=16</math> possible outcomes, since each of the 4 coins can land 2 different ways (heads or tails). There are 2 possibilities for the penny and the dime: either they're both heads or they're both tails. There are also 2 possibilities for the nickel and 2 possibilities for the quarter. So there are <math>2 \times 2 \times 2 = 8</math> successful outcomes, and the probability of success is <math>\dfrac{8}{16} = \boxed{\dfrac{1}{2}}</math>.
 +
{{delete|long title}}

Latest revision as of 12:53, 30 October 2024

There are $2^4=16$ possible outcomes, since each of the 4 coins can land 2 different ways (heads or tails). There are 2 possibilities for the penny and the dime: either they're both heads or they're both tails. There are also 2 possibilities for the nickel and 2 possibilities for the quarter. So there are $2 \times 2 \times 2 = 8$ successful outcomes, and the probability of success is $\dfrac{8}{16} = \boxed{\dfrac{1}{2}}$.

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