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Three people toss a fair coin and the odd one pays for coffee. if the coins all turn up the same, they are tossed again. find the probability that fewer than 4 tosses are needed.

Respuesta :

shinmin

Since the problem doesn’t give a definite question, here are the possible questions and answer:

P(fewer than 4 tosses) 
= P(one toss) + P(two toss) + P(3 toss) 
= (3/4) + (3/4)(1/4) + (3/4)(1/4)^2 
= 0.984375 


Expected value 
= 1 / p 
= 1 / (3/4) 
= 4 / 3 

Variance 
= (1 - p) / p^2 
= (1 - (3/4)) / (3/4)^2 
= (1/4) / (9/16) 
= 4 / 9 

Standard deviation 
= sqrt(Variance) 
= sqrt(4 / 9) 
= 2 / 3

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