Answer:
a) p⁴
b) ( 1 - p)p³
c) 1 - p⁴
Step-by-step explanation:
Data provided in the question:
The coin flips are independent
Probability that coin lands on head, P(H) = p
Thus,
Probability that coin lands on tails, P(T) = ( 1 - p )
Now,
a) P( H,H,H,H ) = P(H) × P(H) × P(H) × P(H)
or
⇒ P( H,H,H,H ) = p × p × p × p = p⁴
b) P( T,H,H,H ) = P(T) × P(H) × P(H) × P(H)
or
⇒ P( T,H,H,H ) = ( 1 - p) × p × p × p = ( 1 - p)p³
c) probability that the pattern T, H, H, H occurs before the pattern H, H, H, H
For obtaining four consecutive Heads (H) we need to get 3 consecutive Heads first.
let we have found our first HHHH sequence somewhere in our sequence. Now, observing the throw before the first H. So we can have a Tails (T), i.e sequence THHH will appear first, or we can have no throw because the first H was our first throw.
(It can't be a H otherwise your HHHH sequence would not be the first one to occur.)
Thus, the only way for HHHH to come first is to start with it, and P(HHHH) = p⁴.
Hence,
P( T, H, H, H occurs before the pattern H, H, H, H) = 1 - p⁴