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 be over this whole fair coin thing. You don't think it wise to accuse your friend 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 A1 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 far or farther from the expected number of heads as 2 is You computed P(AI) question (1b) above. It should be clear that A2 has the same probability as A1. I'm now asking for the probability PO A1 and A2) (d) OK! Well, so if you do get as far or farther from the expected value (as 2 is) twice, ou'd feel much more confident accusing your friend of cheating! Now you're curious ability of getting a result as far or farther Assuming the coin is fair, what's the pro 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

Solution

Back-up Theory

Let X = number of heads turning up in n tosses of a coin. Then, X ~ B(n, p).

If X ~ B(n, p). i.e., X has Binomial Distribution with parameters n and p,

where n = number of trials and p = probability of one success, then

probability mass function (pmf) of X is given by

p(x) = P(X = x) = (nCx)(px)(1 - p)n – x, x = 0, 1, 2, ……. , n …………..(1)

       Expected value of X = E(X) = np.……………………………………..(2)

       If the coin is fair (unbiased), p = ½ ……………………………………(3)

Part (a)

Assuming the coin is fair, [vide (3) and (2) under Back-up Theory], expected value in 10 tosses

= 10 x ½ = 5 ANSWER

Part (b)

Probability of getting 2 heads or farther from expected value = P(X 2)

= 0.0547 ANSWER [this probability is found using Excel Function]

Part (c)

Probability of getting 2 heads or farther from expected value for 2 days in a row

= P(X 2 on 2 days in a row) = P(X 2 on first day) x P(X 2 on first day), because number of heads on first day and number of heads on second day are independent.

So, the required probability = 0.0547 x 0.0547 = 0.0030 ANSWER

Part (d)

Probability of getting 2 heads or farther from expected value for 3 days in a row

= P(X 2 on 3 days in a row) = {P(X 2 on a day)}3 [by the same theory as in (c)]

So, the required probability = (0.0547)3 = 0.0002 ANSWER

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