Consider an ESP experiment in which the subject is asked to identify the suit (s
ID: 3128906 • Letter: C
Question
Consider an ESP experiment in which the subject is asked to identify the suit (spade, heart, diamond, or club) of a card drawn from a standard deck of 52 cards. The subject goes through 4 trials. On each trial, a card is chosen at random from the well-shuffled deck of 52 cards. On each trial, the subject is scored as "correct" or "error," but is not informed of the outcome. Now suppose that our subject does NOT have ESP and so is guessing at random on each trial. The probability of a correct guess is therefore.25 and the probability of an error is.75. In the 4-trial experiment, we will count the number of correct guesses so there are 5 potential outcomes of the experiment: The subject can be correct on 0 or 1 or 2 or 3 or 4 trials. We want to translate our hypothesis that the subject does not have ESP into what our hypothesis predicts with respect to the different possible outcomes of the experiment. That is, if the person doesn't have ESP, what is p(0 correct guess)? p(1 correct?) p(2 correct)? p(3 correct)? and p(4 correct)? To answer this question, we need to analyze the possible outcomes into their component joint events (i.e., sequences of corrects and errors in 4 trials). List the 1 sequence that corresponds to 0 correct in 4 trials and compute "4 things taken 0 at a time" to show that you get 1 sequence. What is the probability of that sequence? So what is the probability of the result "0 correct"? List the sequences giving 1 correct in 4 trials and compute "4 things taken 1 at a time." What is thre probability of each, ordered sequence? So what is the probability of 1 correct in 4 trials? Do the same for 2 correct. Do the same for 3 correct. Do the same for 4 correct. What should the probabilities of "a" through "e" sum up to? Why? Check your work. Look up the binomial distribution for N=4, p=.25. How do the entries compare with your answers?Explanation / Answer
Let I denote incorrect guess
and C denote correct guess
a.) Sequence = { I, I , I , I }
P( sequence) = 0.75 × 0.75 × 0.75 × 0.75 = 0.3164
P( '0' correct) = 0.3163
B.) Sequence = {(C I I I), (I C I I ). , (I I C I ), (I I I C)}
P(C I I I) = 0.25 × 0.75 × 0.75 × 0.75 = 0.1055
P(I C I I ) = 0.75 × 0.25 × 0.75 × 0.75 = 0.1055
P( I I C I) = 0.75 × 0.75 × 0.25 × 0.75 = 0.1055
P( I I I C) = 0.75 × 0.75 × 0.75 × 0.25 = 0.1055
P( 1 correct answer) = 0.1055 + 0.1055 + 0.1055 + 0.1055 = 0.4220
c.) Sequence = { (CCII) , (CIIC), ( CICI) , (ICCI), (ICIC), (IICC)}
P(CCII) = 0.25 × 0.25 × 0.75 × 0.75 = 0.0352
P(CIIC) = 0.25 ×0.75 × 0.75 × 0.25 = 0.0352
Similarly, P(CICI) = P(ICCI) = P( ICIC) = P( IICC)= 0.0352
P( 2 correct) = 0.0352 × 6 = 0.2111
d.) Sequence = {(CCCI) (CICC) (CCIC) (ICCC)}
P(CCCI) = 0.25×0.25×0.25×0.75 = 0.0117
P(CICC) = P(CCIC) = P(ICCC) = 0.0117
P(3 correct) = 4 × 0.0117 = 0.0467
e.) Sequence = {CCCC}
P(CCCC) = 0.25 × 0.25 × 0.25 × 0.25 = 0.0039
P( 4 correct) = 0.0039
f.) The probabilities should sum upto one. This is the sum of all probabilities which is always 1.
0.3163 + 0.2111 + 0.4220 + 0.0467 + 0.0039 = 1