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tossing a coin

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tossing a coin

by sgr21 » Wed Jun 04, 2014 9:37 am
stella and annie are gaming which involves tossing a coin. the game is played in the following manner, stella will toss the coin 4 times in a row. if on any of the coin tosses the side that comes up is the same as the one on the immideately previous throw then stella will lose the game immideately and annie will win.
however if this does not occur on any of the coin tosses then stella will win the game, what is the probability that stella wins the game?

1/2
1/4
1/8
1/12
1/16
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by [email protected] » Wed Jun 04, 2014 9:43 am
Hi sgr21,

This question is small enough in "scale" that you could actually map out (draw) the possibilities on your pad.

Here's how the "math" works though.

I'm going to call the two "sides" of the coin "Heads" and "Tails." For Stella to "win" the game, she has to toss 4 times and the pattern has to be either HTHT or THTH.

Each coin has 2 possible outcomes (Heads or Tails), so 4 tosses have (2)(2)(2)(2) = 16 possible outcomes.

With the two "winning" possibilities listed above, there is a 2/16 = 1/8 probability that Stella will win.

Final Answer: C

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by Brent@GMATPrepNow » Wed Jun 04, 2014 9:46 am
sgr21 wrote:stella and annie are gaming which involves tossing a coin. the game is played in the following manner, stella will toss the coin 4 times in a row. if on any of the coin tosses the side that comes up is the same as the one on the immideately previous throw then stella will lose the game immideately and annie will win.
however if this does not occur on any of the coin tosses then stella will win the game, what is the probability that stella wins the game?

1/2
1/4
1/8
1/12
1/16
For Stella to win, the 1st toss can be ANYTHING (heads or tails), the 2nd toss must be DIFFERENT from the 1st, the 3rd toss must be DIFFERENT from the 2nd and the 4th toss must be DIFFERENT from the 3rd.

So, we can write: P(Stella wins) = P(1st toss is anything AND 2nd toss is different from previous AND 3rd toss is different from previous AND 4th toss is different from previous)
= P(1st toss is anything) x P(2nd toss is different from previous) x P(3rd toss is different from previous) x P(4th toss is different from previous)
= 1 x 1/2 x 1/2 x 1/2
= [spoiler]1/8[/spoiler]
= C

Cheers,
Brent
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by Brent@GMATPrepNow » Wed Jun 04, 2014 10:30 am
sgr21 wrote:stella and annie are gaming which involves tossing a coin. the game is played in the following manner, stella will toss the coin 4 times in a row. if on any of the coin tosses the side that comes up is the same as the one on the immideately previous throw then stella will lose the game immideately and annie will win.
however if this does not occur on any of the coin tosses then stella will win the game, what is the probability that stella wins the game?

1/2
1/4
1/8
1/12
1/16
Here's the tree diagram that Rich noted in his response.
Image
As you can see, there are 16 possible outcomes when we toss the coin 4 times.

When we check the 16 outcomes for situations that are favorable to Stella winning, we see that there are 2 outcomes (one in red and one in green)
Image

So, P(Stella wins) = [spoiler]2/16 = 1/8 = C[/spoiler]

Cheers,
Brent
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