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 14 tossed. What is the probability of being correct 14 or more times by guessing (with 1/2 chance)

Answer:0.09923Step-by-step explanation:We will first use combination which gives total number of ways of selecting r objects out of n[tex]\limits^nC_{r} =\frac{n!}{r! (n-r)!}[/tex]for :20C14 =38760 (using calculator)so assuming sequence of 20 terms, there are 38760 ways of picking the 20 terms.Each toss there are 2 possible outcomes, total possible outcomes for 20 tosses combined are:Total possible outcome = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x2 x 2 x 2 x 2 x 2  = [tex]2^{20}[/tex] =1048576As given, the girl predicts 14 tosses correctly. so, the probability of getting 14 tosses correct is : P(14 Right)= [tex]\frac{20C14}{2^{20} }[/tex] =[tex]\frac{38760}{1048576}[/tex] = 0.03696and So on :P (15 Right) =[tex]\frac{20C15}{2^{20} }[/tex] =[tex]\frac{15504}{1048576}[/tex]=0.01478P (16 Right) =[tex]\frac{20C16}{2^{20} }[/tex] =[tex]\frac{4845}{1048576}[/tex]=0.0462P (17 Right) =[tex]\frac{20C17}{2^{20} }[/tex] =[tex]\frac{1140}{1048576}[/tex]=0.001087P (18 Right) =[tex]\frac{20C18}{2^{20} }[/tex] =[tex]\frac{190}{1048576}[/tex]=0.000181P (19 Right) =[tex]\frac{20C19}{2^{20} }[/tex] =[tex]\frac{20}{1048576}[/tex]=0.000019P (20 Right) =[tex]\frac{20C20}{2^{20}}[/tex] =[tex]\frac{1}{1048576}[/tex]=0.000000953We add all probabilities to get:Probability of getting 14 or more correct is : P (14 or more) = P(14 Right) + P (15 Right) + ..... + P(20 right) = 0.03696 + 0.01478+0.0462+0.001087+0.000181+0.000019+0.000000953 = 0.09923