Let -(H,T)10, put p = 1 /4 and q = 3/4, and define a probability measure P on th

Question

Let -(H,T)10, put p = 1 /4 and q = 3/4, and define a probability measure P on the power set of by P(u)-p#H(w)q#T(w) for each outcome E . Think of as a space of 10 biased but independent coin fiips. Calculate the probabilities of each of the following events. You should use conditional probabilities and the fact that you can look at the probability space as the product of 10 iid Bernouli (1/4) distributions to make your computations simpler. (a) The third flip is the first H (b) The sixth flip is the second H (c) There are more H occurring in the first 4 fiips than in the last 4 flips. (d) The number of H lips is 3 or fewer

Explanation / Answer

1)
third flip is the first H
= (3/4)^2 * (1/4)
= 9/64
b)
sixth flip is the second H
= 5C1 * (3/4)^4 * (1/4) * (1/4)            {first head can be on any flip from 1 to 5 , hence 5 C1}
= 5 * (3/4)^4 * (1/4)^2

c) is little lengtht (X,Y) are number of head of first 4 and last 4 respectively

(X,Y) = x number of heads in first 4 flips

and y number of heads in last 4 flips

these are possible combination possible

(4,3),(4,2),(4,1),(4,0),(3,2),(3,1),(3,0),(2,1),(2,0),(1,0)

P(X = x, Y = y) = P(X =x) P(Y = y)

P(X = x) = = nCx * p^x *(1-p)^(n-x)
= 4Cx* (1/4)^x * (3/4)^(4-x)

similarly P(Y =y) = 4Cy * (1/4)^y * (3/4)^(4-y)

hence

P(X = x, Y = y) =    4Cx* (1/4)^x * (3/4)^(4-x) * 4Cy * (1/4)^y * (3/4)^(4-y)

you have to add all combination of probabilities

d)

P(X = k) = nCk * p^k *(1-p)^(n-k)

= 10Ck * (1/4)^k * (3/4)^(10-k)

P(X < = 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.775875

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