Steve thinks that he has a fair coin with equal probability of landing head or tails. He is ready to change his mind if in a long enough series of flips the coin will land on the same side. Steve decided that "long enough" means the probability of a fair coin landing the same way is less than 1%. What is the shortest series that will allow Steve to declare that the coin is not fair?

A. 7 flips

B. 8 flips

C. 99 flips

D. 101 flips

Maybe C idk

Question

Mcgee

Mathematics

28 April, 04:46

Steve thinks that he has a fair coin with equal probability of landing head or tails. He is ready to change his mind if in a long enough series of flips the coin will land on the same side. Steve decided that "long enough" means the probability of a fair coin landing the same way is less than 1%. What is the shortest series that will allow Steve to declare that the coin is not fair?

A. 7 flips

B. 8 flips

C. 99 flips

D. 101 flips

Maybe C idk

+3

Answers (1)

Know the Answer?

Carina Turner · Answer 1

Answer: A. 7 flips

Step-by-step explanation:

the shortest series that will allow Steve to declare that the coin is not fair = n

The probability of a coin tossed n times to land on the same side is given as;

Pn = 1/2^n

For Pn < 1% = 0.01

1/2^n < 0.01

nLog (1/2) < Log (0.01)

n > Log (0.01) / Log (1/2)

n > 6.64

The shortest n = 7

Therefore, the shortest series that will allow Steve to declare that the coin is not fair

= 7 flips.