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?

Respuesta :

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

A. The probability that the first 4 outcomes are HHHH

Let P(Hi) = p

Then P(HHHH) = p*p*p*p

= p⁴

B. The probability of THHH

P(T) = 1-P(Hi)

= (1-p)p³

c. The probability that THHH would occur before HHHH

HHHH can only occur first given that 1-n does not exist

HHHH is = p⁴

For THHH to occur first it would be

1-P⁴

What is probability?

This is the aspect of mathematics that discusses the likelihood of an event occurring. It tells if a proposition is true or not.

Read more on probability here:

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