1.)
a)
Frequency function for total number of attempts (say k): f(k)

STEP BY STEP:
probability that A will win in first attempt = p1
probability that B will win in 1st attempt = (1-p1)*p2
probability that A will win in 2nd attempt = (1-p1)*(1-p2)*p1
probability that B will win in 2nd attempt = (1-p1)^2*(1-p2)*p2
So on......
probability that A will win in nth attempt=((1-p1)*(1-p2))^(n-1)*p1
probability that B will win in nth attempt=(1-p1)^n*(1-p2)^(n-1)*p2

For total number of attempts = k

f k =2*n + 1 => A will win in it's (n+1)th attempt:
frequency function ( probability that the game will last for k atteamts):
f(k) = ((1-p1)*(1-p2))^n * p1 --Answer

if k =2*n => B will win in it's nth attempt:
f(k) = (1-p1)^n*(1-p2)^(n-1)*p2 --Answer

1.b) probability that player A will win:
p(A) = player A will win in it's 1st attempt OR in 2nd attempt OR 3rd attempt ..... so on
=> p(A) = p1 * (1-p1)*(1-p2)*p1 + ((1-p1)*(1-p2))^2*p1 + ((1-p1)*(1-p2))^3*p1 + ....
p(A) = p1/{1 - ((1-p1)*(1-p2))} --Answer
(Geometric series with 1st term p1 and common ratio (1-p1)*(1-p2))

2.) PLEASE LOOK AT YOUR PROBLEM AND JUST CHECK IF THERE IS ANY TYPO BECAUSE I'M NOT GETTING:
P(X>n+k-1 X>n-1) = P(X>k)
WHAT IS BETWEEN X>n+k-1 AND X>n-1 ?

Here, I'm giving you a brief intro of geometrical random variable.
because,
P(X= n) = q^(n-1) * p
( where p is the probability for event happening and ...

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