Difference between revisions of "2019 AMC 10B Problems/Problem 21"
(Created page with "We know that <math>\dfrac14 + \dfrac14 ^ 2 + \dfrac14 ^ 3 + \cdots = \dfrac13.</math>") |
|||
| Line 1: | Line 1: | ||
| − | We | + | We first want to find out the sequences of coin flips that satisfy this equation. Since Debra sees two tails before two heads, her first flip can't be heads, as that would mean she would either end at tails or see two heads before she sees to tails. Therefore, her first flip must be tails. If you calculate the shortest way she can get two heads in a row and see two tails before she sees two heads, it would be T H T H H, which would be 1/2^5, or 1/32. Following this, she can prolong her coin flipping by adding an extra ( T H), which is an extra 1/4 chance. Since she can do this indefinitely, this is an infinite geometric sequence, which means the answer is (1/32 / (1-1/4) ) or (B) 1/24. (chen1046) |
Revision as of 15:43, 14 February 2019
We first want to find out the sequences of coin flips that satisfy this equation. Since Debra sees two tails before two heads, her first flip can't be heads, as that would mean she would either end at tails or see two heads before she sees to tails. Therefore, her first flip must be tails. If you calculate the shortest way she can get two heads in a row and see two tails before she sees two heads, it would be T H T H H, which would be 1/2^5, or 1/32. Following this, she can prolong her coin flipping by adding an extra ( T H), which is an extra 1/4 chance. Since she can do this indefinitely, this is an infinite geometric sequence, which means the answer is (1/32 / (1-1/4) ) or (B) 1/24. (chen1046)
