Amelia has a coin that lands heads with probability 1/3, and Blaine

Amelia starts so probability of winning first time is 1/3.
If she doesn't win and Blaine doesn't win but Amelia wins on next turn, probability = (2/3)*(3/5)*(1/3) = (2/5)*(1/3)
Probability of Amelia winning next time = (2/5)^2 * (1/3)
and so on..

Probability fo Amelia winning = (1/3) + (2/5)*(1/3) + (2/5)^2 * (1/3) + ...
This is an infinite GP.

Sum = (1/3) / (1 - 2/5) = 5/9

q - p = 4

Answer (D)

For more on GP, check: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html/2012/0 ... gressions/
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A tosses first and A wins so probability of that happening=1/3 therefore q-p= 2
1,3 are relatively prime
If A tosses and A does not win in first toss then probability of winning in 2nd toss= 2/3*3/5*1/3= 2/15 the numbers are relatively prime, but the difference of 13 is not in the options. From here on we see that the denominator will get bigger, so the difference given in options are small and would not fit. We will go with the first q-p value we got which is 2. Hence
IMO B

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Probability of Amelia wins in 1st turn= 1/3
Probability of Amelia wins in 2nd turn= 2/3*3.5*1/3
Probability of Amelia wins in 3rd turn= 2/3*3.5*2/3*3.5*1/3
The probability that Amelia wins= 1/3+2/3*3.5*1/3+2/3*3.5*2/3*3.5*1/3......infinite turns
= (1/3)/{1-(2/3*3/5)}=5/9
p-q=9-5=4

Hey, sumi
1 - is not a prime number

Hey DimasSpb, relatively prime means that only 1 is the number that divide or is common to the pair of given numbers. Example: 14 and 15 are relatively prime. Even though 14 is not a prime number. Hope this helps.

what does relative prime integer means???

Relative primes have HCF=1.

We usually call them co-primes.

e.g. 4 and 5 are co-primes.
8 and 27 are co-primes.
but 10 and 12 are not co-primes because they have a common factor, 2.

Co-primes have only one common factor and that is 1.
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In GP series:
Fixed ratio between any two consecutive terms, denoted by r = a2/a1 = a3/a2 and so on.
Sum of all the terms in a GP series = a1 (1-r^n) / (1-r).

In an infinite GP, as n approaches infinity, ratio r^n approaches zero.
Here, Sum of terms becomes = a1 (1-0) / (1-r) = a1/(1-r)

In this specific question, a1 = 1/3 and r= 2/5 in an infinite GP series
Therefore, Sum of terms= (1/3) / (1-2/5) = (1/3) / (3/5) = 5/9

Hence, difference (denominator - numerator) = 9-5 = 4.

Note:
"p and q are relatively prime positive integers" means the equation p/q is reduced to its smallest form, wherein p and q becomes co-primes. Eg. fraction 30/36 is not in the smallest form as can be further reduced to 5/6.
HCF of 30,36 is 6 whereas HCF of 5,6 is 1 (Co-primes).

another solution --
1- P of winning of B
i.e. in first attempt A looses and then B wins
1-(2/3*2/5)

= 1- 4/15
=11/15
q-p=4

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