M15 #31

John throws a coin until a series of three consecutive heads or three consecutive tails appears. What is the probability that the game will end after the fourth throw?
(A) \frac{1}{16}
(B) \frac{2}{16}
(C) \frac{3}{16}
(D) \frac{4}{16}
(E) \frac{6}{16}

is d ans b....

1/16 + 1/16

2/16

Tricky question!! good one.

this trap for 4/16 (answer D).

TOTAL no. of ways. = 2^4=16

At the first look we may thought all below four combinations are correct.
HTTT
THHH
TTTH -- If this were the case game would have been over after 3 throws.
HHHT-- If this were the case game would have been over after 3 throws.


so only two possiblilites...
ans = 2/16

Nice explanation Suresh. I could not imagine the highlighted part and ended up selecting 4/16 as the answer.

nice explanation, got tricked by this Q.

ohh snapp this is a tricky one. I got the right answer in abt 15 sec but it took me over a minute to realize the trap of 4/16. i went over my answer abt 5 times thinking it was wrong. i really have to spend more time on looking at answer choices before i start cramping the maths.