Let be "X" the amount of offers. Because we have a deal with independent experiments with two outcomes (offer or not), "X" is distributed according to the Binomial distribution "X \\sim B(n,p)" where "n" is the amount of possible 'jobs' and "p" is the probability of success (in our case "p = 0.48"). The PMF function is
"{f_X}(k) = C_n^k{p^k}{(1 - p)^k}"where "C_n^k = {{n!} \\over {k!(n - k)!}}" , CDF function can be calculated as
Part a)
We need to find "P(X \\ge 3)". Let's rewrite it and use the definition of CDF
Let's calculate "{F_X}(2)" (in this case "n = 9")
Thus
Part b)
We need to find such minimal "n" that "P(X \\ge 3) = 0.95" . Actually, we need to solve the equation
but it's impossible to do without using advenced math (such as incomplete beta-fucntions), so we shall just use brute-force.
We shall calculate "{F_X}(2)" for dirrerent "n" (starting with 10) until we have
Let's start with "n = 10"
It doesn't fit, thus continue with "n = 11"
We got it, thus she needs to apply for 11 jobs, and the answer is '2 more'.
Part c)
We need to do the same thing but now we should resolve it for "p"
"{(1 - p)^9} + 9p{(1 - p)^8} + 36{p^2}{(1 - p)^7} = 0.05"
It is the polynomial equation of 9th degree and can't be solved analytically. Thus let's use Newton's method and rewrite our problem as finding a real root of the function
Newton's method says (wery briefly) that we can find the root by iterative process
and with appropriate initial approximation "{p_0}" we can get the solution "p = \\mathop {\\lim }\\limits_{k \\to \\infty } {p_k}" . We can use "p = 0.5" as initial approximation (slightly more then our "p" in the problem). The derivative is "g'(p) = - 252{p^2}{(1 - p)^6}". Calculate
and
and we get for the first approximation
The same procedure for the second approximation give us
We shall stop at this point and approximately says that "p \\approx 0.55" , this is the answer.
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