Question

In a test for ESP (extrasensory perception), the experimenter looks at cards that are hidden from the subject. Each card contains either a star, a circle, a wave, or a square. As the experimenter looks at each of 20 cards in turn, the subject names the shape on the card.

(a) (3pts) Suppose a subject simply guesses the shape on each card, independently from card to card. What is the probability that he guesses correctly the shape on exactly 5 of the 20 cards?

(b) (5pts) What is the probability that a subject correctly guesses at least 10 of the 20 shapes? Use normal approximation to binomial distribution with continuity correction to compute the probability.

(c) (4pts) A standard ESP deck actually contains 25 cards. There are 5 different shapes: a star, a circle, a wave, a cross, or a square; each of which appears on 5 cards. The subject knows that the deck has this makeup (so he knows his 25 answers should also be comprised of 5 stars, 5 circles, 5 waves, 5 crosses, and 5 squares). Is a binomial model still appropriate for the count of correct guesses in one pass through this deck? If so, what are n and p? If not, why not?

Answer 1

n=20

p=1/4 = 0.25

a)

P ( X =    5   ) = C(20,5) * 0.25^5 * (1-0.25)^15 =            0.2023   (answer)

b)

Sample size , n =    20                      
Probability of an event of interest, p =   0.25                      
right tailed                          
X ≥   10                      
                          
Mean = np =    5                      
std dev ,σ=√np(1-p)=   1.9365                      
                          
P(X ≥   10   ) = P(Xnormal ≥   9.5   )          
                          
Z=(Xnormal - µ ) / σ = (   9.5   -   5   ) /   1.9365   =   2.324
                          
=P(Z ≥   2.324   ) =    0.0101              

c)

Binomial Distribution must have same probability of success for each outomce.
Suppose, a star is picked first; p = 5/25= 1/5
Probability of 2nd star card = 4/24 = 1/6 ,
Since 1/5 and 1/6 are differ significantly, we cannot use Binomial