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# Hard Counting and Probability

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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.

P.S. I have already tried my absolute best on this, and I am really, really stuck.

May 9, 2024

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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