karthikpandian19 wrote:When a certain weighted coin is flipped, the probability that it will land on heads is P. What is P?
If the coin is flipped twice in a row, the probability that it will land on tails exactly once is 21/50.
If the coin is flipped twice in a row, the probability that it will land on tails both times is 9/100.
We are given that a weighted coin has probability of landing heads p. Hence probability of landing tales = 1-p. Both p and (1-p) are positive.
With that in mind, let's look at the statements:
If the coin is flipped twice in a row, the probability that it will land on tails exactly once is 21/50.
Exactly once means that we can have two cases:
Tale First * Head Second - probability = (1-p)*p
Head First * Tale Second - probability = p*(1-p).
Hence total probability = 2p*(1-p).
We can stop our calculations at this point. We know that we have a quadratic, and we will definitely get two positive values for p (This is because when we write the quadratic, its indistinguishable whether we are solving for p or 1-p.) This makes it insufficient. I will still solve it though, so that it can be proved to you.
2p(1-p) = 21/50 => p(1-p) = 21/100 = 7/10*3/10. (so p is either 3/10 or 7/10). Insufficient.
If the coin is flipped twice in a row, the probability that it will land on tails both times is 9/100.
Here we only have one case:
Tale First * Tale Second. = (1-p)*(1-p) = (1-p)^2. Again, we don't have to solve any further. equating (1-p)^2 to anything positive will give us only one value for 1-p (because 1-p CANNOT be negative. Then we can find p. Sufficient.
Again, here's the proof: (1-p)^2 = 9/100 => 1-p = 3/10. (As probability can't be -ve).
=> p= 1-3/10 = 7/10. Sufficient.
Hence
B is the correct answer.
I did the whole, "we don't have to calculate" thing, because unlike the question here, they may give you fairly complicated numbers. You don't want to waste time solving the quadratics. A data sufficiency question is, as the name suggests, a question asking, whether we can get a unique answer. If we can determine that without doing calculations, we save valuable time on the exam.
Cheers!
