Layla ha a coin that ha a 60% chance of howing head each time it i flipped. What i the probability that he get more than 3 head?

If there were only 5 coin tosses, the answer would be p5, where p= 0.6.

Now,

18 - 5 + 1 = 14 consecutive sequences of 5 flips contained in 18 flips. So it's tempting to say the answer is 14p5 ≈ 1.08, but it's also wrong. This is actually the number of expected 5-sticks, not the probability of at least one 5-stick. The problem is that since these sequences overlap, the events are not mutually exclusive, so we cannot sum them to get the overall probability. We have to subtract out all the possibilities of generating multiple 5 streaks, but calculating that probability is very complicated.

This post on stack swapping suggests an elegant approach that can be adapted to this problem. Let's define a sequence P(n). where P(n) is the probability of at least one consecutive heads after a total of n rollovers. So the answer of the question is P(18).

Clearly, P(0) = P(1) = P(2) = P(3) = P(4) = 0. This is because he can't get 5 heads in a row unless he flips the coin 5 times. P(5) is easy to compute, it only happens if all 5 flips are heads, so P(5) = p5 = 2433125 = 0.07776 .

Well, the length of the discontinuous part is n - 6, so the probability is

(1 −P(n−6)). The T,H,H,H,H,H part is the probability (1−p)p5 =(0,4)(0,6)5, and since these are independent events, the probability of THHHHH [not consecutive first letters] is (1−P( n −6))( 1−p) p5 .

Putting, All together:

P(n) = P(n−1) + (1−P(n−6))(1−p) p5.

Learn more about coin tosses:

https://brainly.com/question/13384135

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