The Association for Research and Enlightenment (ARE) in Virginia Beach, VA, offe
ID: 3022205 • Letter: T
Question
The Association for Research and Enlightenment (ARE) in Virginia Beach, VA, offers daily demonstrations of a standard technique for testing extrasensory perception (ESP). A "sender" is seated before a box on which one of five symbols (plus, square, star, circle, wave) can be illuminated. A random mechanism selects symbols in such a way that each symbol is equally likely to be illuminated. When a symbol is illuminated, the sender concentrates on it and a "receiver" attempts to identify which symbol has been selected. The receiver indicates a symbol on the receiver's box, which sends a signal to the sender's box that cues it to select and illuminate another symbol. This process of illuminating, sending, and receiving a symbol is repeated 25 times. Each selection of a symbol to be illuminated is independent of the others. The receiver's score (for a set of 25 trials) is the number of symbols that s/he correctly identifies. For the purpose of this exercise, please suppose that ESP does not exist. How many symbols should we expect the receiver to identify correctly? The ARE considers a score of more than 7 matches to be indicative of ESP. What is the probability that the receiver will provide such an indication? The ARE provides all audience members with scoring sheets and invites them to act as receivers. Suppose that, as on August 31, 2002, there are 21 people in attendance: 1 volunteer sender, 1 volunteer receiver, and 19 additional receivers in the audience. What is the probability that at least one of the 20 receivers will attain a score indicative of ESP?Explanation / Answer
a) probability of identifying correctly = 1/5 0.2
Expectation = probability* total trials = 0.2*25 = 5
b) probability of scoring more than 7 = 1- probability of scoring less than or equalt to 7
= 1- (probability of score 0+ probability of score 1 +probability of score 2+ probability of score 3 +probability of score 4+ probability of score 5+probability of score 6+ probability of score 7)
= 1- 0.8^25 + 25C1*0.2*0.8^24+ 25C2*0.2^2*0.8^23+ 25C3*0.2^3*0.8^22+ 25C4*0.2^4*0.8^21+ 25C5*0.2^5*0.8^20+ 25C6*0.2^6*0.8^19+25C7*02.^7*0.8^18)
c) To be indiactive of ESP one needs score of 7
now the trials will happen 19 times instead of 1 so probability will increase by 19 times from part b answwer
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