l. You play a game with your friend everyday that involves flipping a coin 10 ti

Question

l. You play a game with your friend everyday that involves flipping a coin 10 times. Your friend always provides the coin and, you begin to suspect that they are cheating by using weighted coins. In order to test this, you decide to record the number of heads in these 10 flips each day. Some days the coin your friend brings appears to favor heads and other days tails, but your friend always seems to know which it will be that day. (a) What is the expected number of heads when flipping a fair coin 10 times? (b) Suppose one day you get 2 heads in your 10 flips, Assuming the coin is fair, what is the probability of getting a result so far or farther from the expected value? Note: If you're not sure what I mean by "so far or farther", refer back to the explanation in Quiz 1. (c) Hmmm... this probability is low but not crazy low and you've lost friends before over this whole fair coin thing. You don't think it wise to accuse your friend of anything quite yet. Assuming the coin is fair, what is the probability getting a result as far or farther from the expected number of heads (as 2 is) two days in a row? Note: To be clear, consider conducting the experiment of flipping the coin 10 times) twice. Let Au be the event that the first time you flip the coin 10 times you get a result as far or farther from the expected number of heads as 2 is Let A2 be the event that the second time you flip the coin 10 times you get a result as far or farther from the expected number of heads as 2 is. You computed P(A1) in question (1b) above. It should be clear that A2 has the same probability as A1. I'm now asking for the probability PAI and A2). (d) OK! Well, so if you do get as far or farther from the expected value (as 2 is) twice, you'd feel much more confident accusing your friend of cheating! Now you're curious. Assuming the coin is fair, what's the probability of getting a result as far or farther from the expected number of heads (as 2 is) three times in a row? (e) Suppose you repeated this experiment (of flipping the coin 10 times) not twice, not three times, but 20 times! What is the probability of getting a result as far or farther away from the expected value (as 2 is) at least 3 times (out of your 20 experiments)? Hint: You know the probability of getting such a result in any single (10 flip) experiment it's the answer to question (lb). (f) Suppose you steal one of your friends coins. Devise a specific experiment that you could use to test your hypothesis that the coin is unfairly weighted. Specifically, pick some number of times to flip the coin, n, and some criteria for analyzing the results such that there is less than a 0.1% chance of your test leading you to incorrectly accuse your friend of using a weighted coin.

Explanation / Answer

Answering questions E-F as specifically requested

We apply binomial distribution principles

b) Probability of getting 2 heads when a fair coin is flipped 10 times

= 10C2*0.52 *0.58 = 0.0439 or 4.39 %

e) Probability of getting only two heads per 10 flips when repeated 20 times for at least 3 times; P(x>=3)

here we use the same binomial distribution with the outcomes as getting only 2 heads in 10 flips;p and not getting 2 heads in 10 flips;q

from 1b. p = 0.0439, q = 0.9561

Since total probability of all outcomes is 1

P(x>=3) + P(x=2) + P(x=1) + P(x=0) = 1 ,

P(x>=3) = 1-P(x=2) - P(x=1) - P(x=0)

P(x>=3) = 1- 20C2 * 0.04392 * 0.956118 - 20C1 * 0.04391 * 0.956119 - 20C0 * 0.04390 * 0.956120

   = 1 - 0.163 - 0.3742 - 0.4074 = 0.0552 or 5.52%

Probability of getting only two heads per 10 flips when repeated 20 times for at least 3 times = 0.0552 or 5.52%

(f) We need to devise the experiment such that the distribution becomes a normal distribution

Conditions to be satisfied to do that is np and nq be at least 5. where n is number of trials and p and q are probability of getting heads and getting tails respectively, which for a fair sided coin are 0.5 each

np = 5 : n*0.5 = 5

hence n = 5/0.5 = 10

If we conduct 10 flips we can approximate the distribution to a normal distribution

Next we need mean mu and standard deviation

mu = np = 5

sd = square root of npq = 2.51/2 = 1.5811

We set the significance level of 0.1% or 0.001.

By setting the above level we can be sure that we will be confident that our results are correct 99.9% of the times

hence the z score or amount of standard deviations away from mean, where the probability of getting the outcome is less than 0.1% are found out from the normal distribution table.

Since the coin can be biased either ways( towards heads or tails) we split the significance level for both sides , p = 0.0005 and p = 0.9995

z scores = +/- 3.2905

hence no of heads = 3.2905 * sd + mu = 10.2026

hence no of tails also the same = 10.2026

This was not useful as the number of heads/tails can vary from 0-10

We should increase n, let us increase it to 50

hence mu = 25 and sd = 3.5355 at n = 50

here the number of heads at the arived z score = 3.2905 * sd + mu = 36.6336 ~37 heads

conversely 50-37 = 13 tails

and 13 heads at the other end

Hence if we flip the coin 50 times and get less than 13 heads or tails we can be 99.9 % confident that the coin is biased or weighted ( less than 0.1% chance that we are falsely accusing our friend)

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