Two coins are tossed with probability landing at Head not

(1) Suppose we have a coin with 3 Head faces and 1 Tail face, a special biased coin.

Therefore probability of getting head in two tosses = 3/4*3/4= 9/16.

Therefore 1 is sufficient.

If H+T = 1

Would the second statement be

2 (H / H+T) (T/H+T) = 2/9?

Then replacing we get that HT = 9
But still can't do the breakdown

Is this correct?

Cheers
J

HT=(1/9) can be further simplified; as T=(9/h);

substituting the value of T in H+T=1 we will have a quadratic equation in H, which gives two different values of H , hence not sufficient.

I got the first part(statement 1). As they are 100% we can find out that x%+y%=100%, and x%/y%=3 Gives us 75%/25%=3.

The second part I just "panicked" and couldn't solve. Will try and read up on probability.

Hi,

Could you please explain the second statement a bit more clearly ? I mean why 2 is not sufficient?

Thanx in advance.

Two coins are tossed with probability landing at Head not equal to 0.5. What is the probability of getting two Heads out of two tosses?

Say the probability of heads is p and the probability of tails is 1-p.

(1) The probability of getting head is 3 times that of getting tails --> p=3(1-p) --> p=3/4 --> the probability of getting two Heads out of two tosses = 3/4*3/4. Sufficient.

(2) Tossed two times, the probability of getting one Head and one Tail is 2/9 --> P(HT)=2*p(1-p)=2/9 (we multiply by to because one Head and one Tail can occur in two ways HT and TH). We get two value for p. Not sufficient.

Answer: A.

P.S. This is a poor quality question, since the values of p differ from (1) and (2), while on the GMAT, two data sufficiency statements always provide TRUE information and these statements never contradict each other or the stem.