A weighted coin has a probability p of showing heads. Is the value of

Question

A weighted coin has a probability p of showing heads. Is the value of p > 0.2?

(1) Successive flips are independent, and the probability of getting at least one head in two flips is greater than 0.5.

(2) The probability of getting tail on a flip of coin is less than 0.8.

GMATinsight · Answer

Question : is p > 0.2? Statement 1 : Probability of two tails < 0.5 i.e. (1-p)*(1-p) < 0.5 i.e. (1-p)^2 < 0.5 i.e. (1-p) < 0.707 i.e. p > 0.293 SUFFICIENT Statement 2 : (1-p) < 0.8 i.e. p > 0.2 SUFFICIENT Answer: option D

elPatron434 · Answer

Hi GMATinsight , I'm getting a different value of p. Could you please help me with where I'm going wrong?

2C1*p(1-p) + p^2 > 0.5
2p - 2p^2 + p^2 > 0.5
2p - p^2 - 0.5 > 0
4p - 2p^2 -1 > 0
2p^2 -4p +1 < 0

b^2 - 4ac = 16-8 = \sqrt{8}
Roots are 1+0.5\sqrt{2} or 1-0.5\sqrt{2}

So p< 1-0.5\sqrt{2}

IanStewart · Answer

Assuming this non-standard coin has tails on the other side, we can rephrase the question: is the probability of tails less than 0.8? Say that probability is q, so we want to know if q < 0.8.

Statement 1 tells us the probability of getting tails twice in a row is less than 1/2. So q^2 < 1/2, and q < 1/√2. Since 1/√2 is very close to 0.7, we can be sure q < 0.8, and Statement 1 is sufficient.

Statement 2 tells us exactly what we wanted to know, so the answer is D.

IanStewart · Answer

I don't think that's the best way to approach the problem, but your work is entirely correct, except for the very last line. You have this inequality 2p^2 -4p +1 < 0 You found the two roots -- just for convenience, let's call them "r" and "s", where r = 1 + (√2/2), and s = 1 - (√2/2). So you can factor (p - r)(p - s) < 0 That inequality means one factor is positive, the other is negative. Since r is bigger than s, p-r is smaller than p-s (when we subtract a larger value from p, we get a smaller result). So if one of p-r or p-s is negative, it must be p-r that is negative, since it's smaller than p-s. So p - r < 0, and p < r, and p - s > 0, and p > s. So the correct conclusion to draw from your quadratic inequality is the opposite of the one you reached -- it should be: 1 - (√2/2) < p < 1 + (√2/2) Of course the righthand part of this inequality is useless, because we already know more than that; of course, p < 1. It's the lefthand part that is useful: 1 - (√2/2) < p, which, to one decimal place, means roughly that p > 0.3.

elPatron434 · Answer

Aah I missed this! Thanks a ton  IanStewart

Posted from my mobile device

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A weighted coin has a probability p of showing heads. Is the value of

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A weighted coin has a probability p of showing heads. Is the value of

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