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QuestionWhat are the odds of a perfect bracket if a user predicts 66.7%66.7\% of games correctly? A. 1 in 120.2 billion B. 1 in 66.7 C. 1 in 1,202 D. 1 in 120

Studdy Solution

STEP 1

Assumptions1. The average user correctly predicts the outcome of approximately66.7% of the games in their bracket. . There are63 games in a NCAA bracket (since there are64 teams and each game eliminates one team until there is a champion).
3. Each game is an independent event, meaning the outcome of one game does not affect the outcome of another.

STEP 2

First, we need to convert the percentage of correctly predicted games to a decimal.
Correct prediction rate=66.7%=0.667Correct\, prediction\, rate =66.7\% =0.667

STEP 3

The probability of predicting each game correctly is0.667. Since each game is an independent event, the probability of predicting all games correctly (a perfect bracket) is the product of the probabilities of each individual game.
Probability of perfect bracket=(Correct prediction rate)Number of gamesProbability\, of\, perfect\, bracket = (Correct\, prediction\, rate)^{Number\, of\, games}

STEP 4

Plug in the values for the correct prediction rate and the number of games to calculate the probability of a perfect bracket.
Probability of perfect bracket=(0.667)63Probability\, of\, perfect\, bracket = (0.667)^{63}

STEP 5

Calculate the probability of a perfect bracket.
Probability of perfect bracket=(0.667)63≈1.15×10−20Probability\, of\, perfect\, bracket = (0.667)^{63} \approx1.15 \times10^{-20}

STEP 6

dds are typically expressed as a ratio of the number of successful outcomes to the number of unsuccessful outcomes. In this case, the successful outcome is getting a perfect bracket, and the unsuccessful outcome is not getting a perfect bracket. Since the probability of a perfect bracket is extremely small, the odds are approximately1 to the reciprocal of the probability.
dds of perfect bracket=11Probability of perfect bracketdds\, of\, perfect\, bracket =1 \frac{1}{Probability\, of\, perfect\, bracket}

STEP 7

Plug in the value for the probability of a perfect bracket to calculate the odds.
dds of perfect bracket=111.15×10−20dds\, of\, perfect\, bracket =1 \frac{1}{1.15 \times10^{-20}}

STEP 8

Calculate the odds of a perfect bracket.
dds of perfect bracket=111.15×10−20≈18.7×1019dds\, of\, perfect\, bracket =1 \frac{1}{1.15 \times10^{-20}} \approx18.7 \times10^{19}This is approximately1 in87 quintillion, which is much larger than any of the given options. Therefore, none of the options A, B, C, or D are correct. The odds of a user getting a perfect bracket, given that they correctly predict66.7% of the games, are approximately1 in87 quintillion.

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