solve (e) (f) (g) Problem(4) (a) A biased coin is tossed, and it is assumed that

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solve (e) (f) (g)

Problem(4) (a) A biased coin is tossed, and it is assumed that the chance of gettinga head, H, is . (Thus the chance of getting a tail, T, is Consider a random experiment of throwing the coin 5 times. Let S denote the sample space. Describe the elements in S (b) Let X be the random variable that corresponds to the number of the heads coming up in the five times of toss. What are the values that the random variable X takes? (c) Find the probability that there is two tails and three heads, that is, P(x- 3) (d) Find the probability that at least one of them lands head, that is, P(X 21 Suppose that for each toss that comes up heads we win $10, but for each toss that comes up tails we lose $8. Clearly, a quantity of interest in this situation is our total wining. Let Y denote this quantity, what are the values that the random variable Y takes? (f) Find P(Y = 14) (g) Find P(Y s 10)

Explanation / Answer

e) Here we are given that the biased coin is tossed 5 times. Also we are given that the head comes with a probability of 0.4 and tails comes with a probability of 0.6

Now for different number of heads / tails, we get different values of Y. Here the values of Y along with its probability could be computed as:

Therefore the different values of Y possible here are: Y = { -40, -22, -4, 14, 32, 50 }

f) P(Y = 14 ) = 0.2304 ( from the above table )

Note that the probabilities in the table are computed using the binomial probability function

g) P( Y < =10 ) = P(Y=-40 ) + P(Y = - 22) + P(Y = -4 ) = 0.07776 + 0.2592 + 0.3456 = 0.68256

Therefore the required probability here is 0.68256

Number of heads Number of Tails Probability Y 0 5 0.65 = 0.07776 0 - 8*5 = -40 1 4 5*0.4*0.64 = 0.2592 10 - 8*4 = -22 2 3 10*0.420.63 = 0.3456 10*2 - 8*3 = -4 3 2 10*0.430.62 = 0.2304 10*3 - 8*2 = 14 4 1 5*0.6*0.44 = 0.0768 10*4 - 8 = 32 5 0 0.45 = 0.01024 10*5 = 50

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