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The probability of a 3-person jury making a correct decision is the union of two situations (here, Juror 1 and 2 have probability P of being right, while Juror 3 has a 50% probability of being right): | The probability of a 3-person jury making a correct decision is the union of two situations (here, Juror 1 and 2 have probability P of being right, while Juror 3 has a 50% probability of being right): | ||
<pre> | |||
P_{3 correct} = ( P_{juror 1 correct} and P_{juror 3 correct} ) or (P_{juror 2 correct} and P_{juror 3 correct}) | P_{3 correct} = ( P_{juror 1 correct} and P_{juror 3 correct} ) or (P_{juror 2 correct} and P_{juror 3 correct}) | ||
</pre> | |||
Now we know P_{juror 3 correct} is 0.5, we know that "and" translates to multiplication, we know that "or" translates to addition, so: | Now we know P_{juror 3 correct} is 0.5, we know that "and" translates to multiplication, we know that "or" translates to addition, so: | ||
<pre> | |||
P_{3 correct} = 0.5*P + 0.5*P = P | P_{3 correct} = 0.5*P + 0.5*P = P | ||
</pre> | |||
Meanwhile, the 1 person jury has a probability of making a correct decision of: | Meanwhile, the 1 person jury has a probability of making a correct decision of: | ||
<pre> | |||
P_{1 correct} = P | P_{1 correct} = P | ||
</pre> | |||
From this, we can see that the two juries have an equal probability of being correct. | From this, we can see that the two juries have an equal probability of being correct. | ||
}} | }} | ||
Revision as of 22:20, 16 November 2019
Friday Morning Math Problem
The Flippant Juror
A 3-person jury has 2 members, each of whom has an independent probability P of making a correct decision. The third juror flips a coin. (Majority rules.)
A 1-person jury has 1 member who has a probability P of making the correct decision.
Which jury has a better probability of making the correct decision?
| Solution |
|---|
|
The probability of a 3-person jury making a correct decision is the union of two situations (here, Juror 1 and 2 have probability P of being right, while Juror 3 has a 50% probability of being right):
P_{3 correct} = ( P_{juror 1 correct} and P_{juror 3 correct} ) or (P_{juror 2 correct} and P_{juror 3 correct})
Now we know P_{juror 3 correct} is 0.5, we know that "and" translates to multiplication, we know that "or" translates to addition, so: P_{3 correct} = 0.5*P + 0.5*P = P
Meanwhile, the 1 person jury has a probability of making a correct decision of: P_{1 correct} = P
From this, we can see that the two juries have an equal probability of being correct. |
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Friday Morning Math Problems weekly math problems
Spider Socks and Shoes FMM1 Sums of Powers of 2 FMM2 Fifty Coins FMM3 The Zeta Monogram FMM4 The Cthulhus Monogram FMM4B Multiplication Logic FMM5 The Termite and the Cube FMM6 Sharing Dump Trucks FMM7 The Flippant Juror FMM8 Bus Routes FMM9 A Robust Bus System FMM9B Square-Free Sequence FMM10 Inferring Rule from Sequence FMM11 Checkerboard Color Schemes FMM12 One-Handed Chords FMM13 First Ace FMM14 Which Color Cab FMM15 Petersburg Paradox Revisited FMM16 A Binomial Challenge FMM17 A Radical Sum FMM18 Memorable Phone Numbers FMM19 Arrange in Order FMM20 A Pair of Dice Games: FMM21
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