MATH SOLVE

3 months ago

Q:
# Independent flips of a coin that lands on heads with probability p are made. What is the probability that the first four outcomes are (a) H,H,H,H? (b) T,H,H,H? 3 (c) What is the probability that the pattern T, H, H, H occurs before the pattern H, H, H, H? Hint for part (c): How can the pattern H, H, H, H occur first?

Accepted Solution

A:

Answer:a) p⁴b) ( 1 - p)p³c) 1 - p⁴Step-by-step explanation:Data provided in the question:The coin flips are independentProbability that coin lands on head, P(H) = pThus,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 = p⁴b) 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)p³c) probability that the pattern T, H, H, H occurs before the pattern H, H, H, HFor 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