To solve this problem we need the knowledge of binomial probability
Our first aim is to find the value of p using the given conditions
Probability of getting 4 head 5 tail can be found out using Binomial probability
C(9,4)p⁴(1-p)⁵
Probability of getting 5 head 4 tail
Is C(9,5)p⁵(1-p)⁴
C(9,4)p⁴(1-p)⁵=1/3[C(9,5)p⁵(1-p)⁴]
3p⁴(1-p)⁵=p⁵(1-p)⁴
3(1-p)=p
p=3/4
Probability of getting two head is
P(2H)=3/4*3/4=9/16
Now there are infinite number of cases possible that is the person who is playing first can win in the first chance can win in the second chance can win in the third chance and so on
Case 1
The person wins in the 1st chance P=9/16
Case 2
The person who wins in the second chance means that the first person losses the first chance and the second person also losses his first chance
and there are infinite cases
P(losing)=1-9/16=7/16
Total probability =7/16 × 7/16 × 9/16=(7/16)²9/16
Total probability
=9/16+(7/16)²9/16(7/16)⁴9/16..∞
=9/16(1+(7/16)²+(7/16)⁴...∞)
=9/16 * 256/207
=16/23
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