2013 AMC 8 Problem 8

Attempt Problem 8 of the 2013 AMC 8 below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

8.

A fair coin is tossed 33 times. What is the probability of at least two consecutive heads?
18\dfrac{1}{8}
14\dfrac{1}{4}
38\dfrac{3}{8}
12\dfrac{1}{2}
34\dfrac{3}{4}
Answer: C
Concepts:basic probabilitysystematic listing
Small hint:
List the 88 possible outcomes of three coin tosses.
Big hint:
Count only outcomes containing HHHH.
Video solution:
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Written solution:
The outcomes with at least two consecutive heads are HHTHHT, THHTHH, and HHHHHH. There are 23=82^3=8 equally likely outcomes from three fair coin tosses, so the probability is 38\dfrac{3}{8}.
Thus, C is the correct answer.
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