Let E1 be the event of selecting a fair coin and E2 be the event of selecting the two-headed coin.
P(E1)=N+1N
P(E2)=N+11
Let H be the event of getting a Head.
P(H∣E1)=21
P(H∣E2)=1
Using the theorem of total probability:
P(H)=P(E1)P(H∣E1)+P(E2)P(H∣E2)
P(H)=(N+1N)(21)+(N+11)(1)
P(H)=2(N+1)N+2
Given P(H)=169:
2(N+1)N+2=169
N+1N+2=89
8N+16=9N+9
N=7
Answer: 7