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A coin is tossed 20 times. A​ person, who claims to have extrasensory​ perception, is asked to predict the outcome of each flip in advance. She predicts correctly on 11 tosses. What is the probability of being correct 11 or more times by​ guessing? Does this probability seem to verify her​ claim?

Respuesta :

1. 
[tex]C(n, r)= \frac{n!}{r!(n-r!)} [/tex] gives the total number of ways of selecting r objects out of n.

for example 
[tex]C(20, 11)= \frac{20!}{11!(20-11!)}=\frac{20!}{11!9!}= \frac{20*19*18*17*16*15*14*13*12*11!}{11!*9!}=[/tex]

[tex] \frac{20*19*18*17*16*15*14*13*12}{9!}=362,880[/tex], using a calculator.


so assume a sequence of 20 terms, there are C(20, 11) ways of picking exactly 11 of these terms to represent the Right Prediction, for example:

 R R W R W W R W R R R W W R W R W W R R 

(so first time Right guess, second time Right, third Wrong, fourth Right and so on )
or 


 R  W R W R W W R R W W R W R R W R R R W

The total number of possible outcomes is

2*2*......2*2 = [tex] 2^{20} [/tex]= 1,048,576

(we have 20 2's, because each time we can get it Right, or Wrong)

Probability of getting exactly 11 right is [tex] \frac{362,880}{1,048,576}=0,346[/tex]

Similarly the probability of being right exaclty 12, 13, ...20 times is:


[tex]p(12R)= \frac{C(20, 12)}{1,048,576}= \frac{ \frac{20!}{12!8!} }{1,048,576}=[/tex]=0.12

[tex]p(13R)= \frac{C(20, 13)}{1,048,576}= \frac{ \frac{20!}{13!7!} }{1,048,576}=[/tex]=0.074

[tex]p(14R)= \frac{C(20, 14)}{1,048,576}= \frac{ \frac{20!}{14!6!} }{1,048,576}=[/tex]=0.037

[tex]p(15R)= \frac{C(20, 15)}{1,048,576}= \frac{ \frac{20!}{15!5!} }{1,048,576}=[/tex]=0.0148

[tex]p(16R)= \frac{C(20, 16)}{1,048,576}= \frac{ \frac{20!}{16!4!} }{1,048,576}=[/tex]=0.0046

[tex]p(17R)= \frac{C(20, 17)}{1,048,576}= \frac{ \frac{20!}{17!3!} }{1,048,576}=[/tex]=0.001

[tex]p(18R)= \frac{C(20, 18)}{1,048,576}= \frac{ \frac{20!}{18!2!} }{1,048,576}=[/tex]=0.0002

[tex]p(19R)= \frac{C(20, 19)}{1,048,576}= \frac{ \frac{20!}{19!1!} }{1,048,576}=[/tex]=0.00002

[tex]p(20R)= \frac{C(20, 20)}{1,048,576}= \frac{ 1 }{1,048,576}=[/tex]=0.000001

Probability of getting 11 or more right, is P(11)+P(12)+...+P(20)=0.598

Predicting 11 or more times correctly has a probability of 0.598

so any one, with any random prediction has a chance of predicting 11 times or more, correctly, a little more than half of the times.

The woman predicted 11 correct times, which has a probability of 0,346 which is not very small probability. This is not even close to some unusual prediction.

An unusual prediction would be predicting for example 19 times Right, which has a very small probability, just 0.00002

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