Difference between revisions of "2019 AMC 10A Problems/Problem 22"
Pratimakarpe (talk | contribs) (→Solution 2: Commenting out solution 2 as it is incorrect. The probability of a difference of exactly 1/2 is 0, not 1/8th, and the probability of <1/2 and >1/2 is not symmetric, as shown in Solution 1) |
Sevenoptimus (talk | contribs) m (Fixed minor formatting issues) |
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Each case has a <math>\frac{1}{4}</math> chance of occurring. | Each case has a <math>\frac{1}{4}</math> chance of occurring. | ||
− | For | + | For Case 1, we need <math>x</math> and <math>y</math> to be different. Therefore, the probability for success in Case 1 is <math>\frac{1}{2}</math>. |
− | For | + | For Case 2, if <math>x</math> is 0, we need <math>y</math> to be in the interval <math>(\frac{1}{2}, 1]</math>. If <math>x</math> is 1, we need <math>y</math> to be in the interval <math>[0, \frac{1}{2})</math>. Regardless of what <math>x</math> is, the probability for success for Case 2 is <math>\frac{1}{2}</math>. |
− | By symmetry, | + | By symmetry, Case 3 has the same success rate as Case 2. |
− | For | + | For Case 4, we must use geometric probability because there are an infinite number of pairs <math>(x, y)</math> that can be selected, whether they satisfy the inequality or not. Graphing <math>|x-y| > \tfrac{1}{2}</math> gives us the following picture where the shaded area is the set of all the points that fulfill the inequality: |
<asy> | <asy> | ||
Line 37: | Line 37: | ||
Adding up the success rates from each case, we get: | Adding up the success rates from each case, we get: | ||
− | <math>(\frac{1}{4}) \cdot (\frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{4}) = \boxed{\textbf{(B) }\frac{7}{16}}</math>. | + | <math>\left(\frac{1}{4}\right) \cdot \left(\frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{4}\right) = \boxed{\textbf{(B) }\frac{7}{16}}</math>. |
<!--==Solution 2== | <!--==Solution 2== |
Revision as of 21:42, 17 February 2019
- The following problem is from both the 2019 AMC 10A #22 and 2019 AMC 12A #20, so both problems redirect to this page.
Problem
Real numbers between 0 and 1, inclusive, are chosen in the following manner. A fair coin is flipped. If it lands heads, then it is flipped again and the chosen number is 0 if the second flip is heads and 1 if the second flip is tails. On the other hand, if the first coin flip is tails, then the number is chosen uniformly at random from the closed interval . Two random numbers
and
are chosen independently in this manner. What is the probability that
?
Solution 1
There are several cases depending on what the first coin flip is when determining and what the first coin flip is when determining
.
The four cases are:
is either
or
and
is either
or
is either
or
and
is chosen from the interval
is is chosen from the interval
and
is either
or
is is chosen from the interval
and
is chosen from the interval
Each case has a chance of occurring.
For Case 1, we need and
to be different. Therefore, the probability for success in Case 1 is
.
For Case 2, if is 0, we need
to be in the interval
. If
is 1, we need
to be in the interval
. Regardless of what
is, the probability for success for Case 2 is
.
By symmetry, Case 3 has the same success rate as Case 2.
For Case 4, we must use geometric probability because there are an infinite number of pairs that can be selected, whether they satisfy the inequality or not. Graphing
gives us the following picture where the shaded area is the set of all the points that fulfill the inequality:
The shaded area is , which means the probability for success for case 4 is
.
Adding up the success rates from each case, we get:
.
See Also
2019 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2019 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.