Applications of Probability?

23 A student needs to know details of a class assignment that is due the next day and decides to call fellow class members for this information. She believes that for any particular call the probability of obtaining the necessary information is 0.40. She decides to continue calling class members until the information is obtained. Let the random variable X denote the number of calls needed to obtain the information.





a. Find the probability function of X.


b. Find the cumulative probability function of X.


c. Find the probability that at least three calls are required.

Applications of Probability?
a. The probability function of a random variable X that has a geometric distribution is





f(x) =


{ p(1-p)^(x-1), x = 1, 2, 3, ...


{ 0, otherwise.





Since X is the number of trials it takes to get the first success (a student with an answer), then X has a geometric distribution, where the probability of success is 0.4. The probability function of X is





f(x) =


{ 0.4*0.6^(x-1), x = 1, 2, 3, ....


{ 0, otherwise.





b. The cumulative distribution function of the geometric is





F(x) =


{1 - (1-p)^[x], x %26gt; 0


{ 0, otherwise,





where [x] is the floor function of x, which is the largest integer less than x. ([2.6] = 2, [11.9] = 11, and so on)





So in this case,





F(x) =


{ 1 - 0.6^[x], x %26gt; 0


{ 0, otherwise.





c. You are looking for P(X %26gt;= 3). However, that is an infinite sum, so it would be easier to find the complement of X %26gt;= 3 and subtract from 1.





P(X%26gt;= 3) = 1 - P(X %26lt; 3)





= 1 - {P(X = 1) + P(X = 2)}





= 1 - {0.4(0.6^0) + 0.4(0.6^1)}





= 1 - 0.4 - 0.24





= 0.36





edit: fernando, you need to redo parts a. and c.
Reply:This can be modeled by Bernoulli trials with probability of success p = 0.4. Therefore:


a. P[X = k] = p*(1 - p)^(k-1) = (2/3)*0.6^k


b. P[X %26lt;= k] = sum(i = 1 to k) P[X=i] = 1 - (1 - p)^k = 1 - 0.6^k.


c. P[X = 3] = (2/3)*0.6^3 = 0.144.

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