1. You have 100 coins, and 99 of them are fair (equal probability of heads or ta
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1. You have 100 coins, and 99 of them are fair (equal probability of heads or tails). One of them is weighted and has a 90% probability of landing on heads. You randomly choose one of the 100 coins. Find the probability that it is a weighted coin, under the following scenarios: (Hint: if your calculator can't compute 100!, R can, just type factorial(100)) a) You flip it 10 times and lands on heads 10 times (b) You flip it 10 times and it lands on heads 9 times (c) You flip it 20 times and it lands on heads 18 times (d) You flip it 100 times and it lands on heads 77 times 2. A professor gives students a pop quiz with 5 true or false questions. Eighty percent of the students are well-prepared for the pop quiz, but twenty percent are not. Students who are prepared have a 90% chance of answering each question correctly, but the students who are unprepared simply randomly guess and have a 50% chance. Find the probability that a student was well-prepared under the following scenarios: (a) Answered 1 correctly (b) Answered 2 correctly (c) Answered 3 correctly (d) Answered 4 correctly 3. A fair coin is flipped 100 times. Your friend proposes a bet where if it lands on heads 49, 50, or 51 times, you win S4, but if it lands on heads more than 51 times or less than 49 times, you lose S1. Should you take the bet? 4. Two candidates face each other in an election. The Democratic candidate is sup- ported by 58% of the population, and the Republican candidate is supported by 42%. In other words, if you randomly chose a voter and asked them who they plan to vote for, there would be a 58% chance they would say they support the Democratic candidate. Suppose you run a poll of 8 people. What is the probability that less than half of them (3 or fewer) would support the Democratic candidate?Explanation / Answer
(a) Here we have 100 coins, 99 of them are reasonable and one of them is weightd and 90% probability of arriving on heads.
We flip it 10 times and terrains on heads 10 times
Pr(Weighted coin l flip it 10 times and grounds on heads 10 times )
probability of any coin arrival takes 10 times off of 10 times
Pr(landing 10 time head on 10 times) = Pr(Normal coin) * Pr(Normal Coin landing 10 heads on 10 times) + Pr(Weighted Coin) * Pr(weighted Coin landing 10 heads on 10 times) = 99/100 * (0.5)10 + 1/100 * (0.9)10
Pr(Weighted Coin) = 1/100 * (0.9)10/[99/100 * (0.5)10 + 1/100 * (0.9)10] = 0.7829
(b) Pr(Flip 10 times and terrains 9 timess) = Pr(Normal coin) * Pr(Normal Coin landing 9 heads on 10 times) + Pr(Weighted Coin) * Pr(weighted Coin landing 9 heads on 10 times)
= 99/100 * 10C9 (0.5)10 + 1/100 * 10C9 (0.9)9 (0.1)
Pr(Weighted coin) = [1/100 * 10C9 (0.9)9 (0.1) ]/[99/100 * 10C9 (0.5)10 + 1/100 * 10C9 (0.9)9 (0.1) ] = 0.2861
(c) Here flip it 20 times and terrains on heads 18 times
= Pr(Normal coin) * Pr(Normal Coin landing 18 heads on 20 times) + Pr(Weighted Coin) * Pr(weighted Coin landing 18 heads on 20 times)
= 99/100 * 20C18 (0.5)20 + 1/100 * 20C18 (0.9)18 (0.1)2
Pr(Weighted coin) = [1/100 * 20C18 (0.9)18 (0.1)2 ]/[99/100 * 20C18 (0.5)20 + 1/100 * 20C18 (0.9)18 (0.1)2 ] = 0.9408
(d) Here flip it 100 times and it arrives on 77 times
Pr(Head on 77 times out of 100) =
Pr(Normal coin) * Pr(Normal Coin landing 77 heads on 100 times) + Pr(Weighted Coin) * Pr(weighted Coin landing 77 heads on 100 times)
= 99/100 * 100C77 (0.5)100 + 1/100 * 100C77(0.9)77 (0.1)23
Pr(Weighted coin) = [1/100 * 100C77(0.9)77 (0.1)23 ]/[99/100 * 100C77 (0.5)100 + 1/100 * 100C77(0.9)77 (0.1)23 ] = 0.9746
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