+27 A fair coin is tossed repeatedly until either heads comes up three times in a row or tails comes up three times in a row. What is the probability that the coin will be tossed more than 10 times? Express your answer as a common fraction.
Thanks in advance!
P.S. I have already tried my absolute best on this, and I am really, really stuck.
+1439 This problem can be solved by considering the complementary probability (the probability of the opposite event happening) and subtracting it from 1.
Event: The coin is tossed no more than 10 times (either heads or tails appear three times in a row within 10 tosses).
Complementary Event: The coin is tossed more than 10 times (neither heads nor tails appear three times in a row within 10 tosses).
Probability of Complementary Event:
There are two possibilities for the complementary event:
HHH appears within the first 10 tosses: There are 10 possibilities for the location of the first H (tosses 1 through 10). For each of these possibilities, there are two remaining tosses that must be Hs (total of 2 * 10 = 20 outcomes).
TTT appears within the first 10 tosses: Similar to case 1, there are 20 outcomes (10 possibilities for the first T and two remaining tosses must be Ts).
The total number of successful outcomes for the complementary event is 20 (HHH) + 20 (TTT) = 40.
Since each outcome (sequence of tosses) has an equal probability (either H or T on each toss), the probability of any specific outcome is 1/2 raised to the power of the number of tosses (2^-# of tosses).
The total number of possible outcomes for 10 tosses is 2^10 (either H or T on each toss).
Therefore, the probability of the complementary event is:
Probability (complementary event) = (Favorable outcomes) / (Total possible outcomes) Probability (complementary event) = 40 / (2^10)
Probability of the Event:
The probability of the event we're interested in (tossing more than 10 times) is the complement of the complementary event.
Probability (event) = 1 - Probability (complementary event)
Since all outcomes are equally likely, the sum of the probabilities of all possible events must be 1.
Probability (event) = 1 - 40 / (2^10)
Simplifying the Fraction:
We can rewrite 2^10 as 1024.
Probability (event) = 1 - (40 / 1024)
Probability (event) = 1 - (5 / 128)
Probability (event) = 123/128
Therefore, the answer is 123/128.
