Question

An MBA graduate is applying for nine jobs, and believes that she has in each of the nine cases a constant and independent 0.48 probability of getting an offer.
a. What is the probability that she will have at least three offers?
b. If she wants to be 95% confident of having at least three offers, how many more jobs should she apply for? (Assume each of these additional applications will also have the same probability of success.)
c. If there are no more than the original nine jobs that she can apply for, what value of probability of success would give her 95% confidence of at least three offers?

Answer 1

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

$${F_X}(k) = \sum\limits_{i = 0}^k {C_n^i{p^i}{{(1 - p)}^{n - i}}}$$

Part a)

We need to find $P(X \ge 3)$. Let's rewrite it and use the definition of CDF

$$P(X \ge 3) = 1 - P(X \le 2) = 1 - {F_X}(2)$$

Let's calculate ${F_X}(2)$ (in this case $n = 9$)

$${F_X}(2) = {{9!} \over {0!9!}}{0.48^0} \cdot {0.52^9} + {{9!} \over {1!8!}}{0.48^1}{0.52^8} + {{9!} \over {2!7!}}{0.48^2}{0.52^7} \approx 0.11115$$

Thus

$$P(X \ge 3) \approx 1 - 0.11115 \approx 0.88885$$

Part b)

We need to find such minimal $n$ that $P(X \ge 3) = 0.95$ . Actually, we need to solve the equation

$$1 - \sum\limits_{i = 0}^2 {C_n^i{p^i}{{(1 - p)}^{n - i}}} \ge 0.95$$

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

$${F_X}(2) \le 1 - 0.95 \le 0.05$$

Let's start with $n = 10$

$${F_X}(2) = {{10!} \over {0!10!}}{0.48^0} \cdot {0.52^{10}} + {{10!} \over {1!9!}}{0.48^1}{0.52^9} + {{10!} \over {2!8!}}{0.48^2}{0.52^8} \approx 0.07022$$

It doesn't fit, thus continue with $n = 11$

$${F_X}(2) = {{11!} \over {0!11!}}{0.48^0} \cdot {0.52^{11}} + {{11!} \over {1!10!}}{0.48^1}{0.52^{10}} + {{11!} \over {2!9!}}{0.48^2}{0.52^9} \approx 0.04361$$

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 $p$

$${F_X}(2) = \sum\limits_{i = 0}^2 {C_n^i{p^i}{{(1 - p)}^{n - i}}} = 0.05$$ $${{9!} \over {0!9!}}{p^0}{(1 - p)^9} + {{9!} \over {1!8!}}{p^1}{(1 - p)^8} + {{9!} \over {1!7!}}{p^2}{(1 - p)^7} = 0.05$$

$${(1 - p)^9} + 9p{(1 - p)^8} + 36{p^2}{(1 - p)^7} = 0.05$$

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

$$g(p) = {(1 - p)^9} + 9p{(1 - p)^8} + 36{p^2}{(1 - p)^7} - 0.05$$

Newton's method says (wery briefly) that we can find the root by iterative process

$${p_{k + 1}} = {p_k} - {{g({p_k})} \over {g'({p_k})}}$$

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

$$g({p_0}) = {(1 - 0.5)^9} + 9 \cdot 0.5{(1 - 0.5)^8} + {360.5^2}{(1 - 0.5)^7} \approx 0.039844$$

and

$$g'({p_0}) = - 252 \cdot {0.5^2}{(1 - 0.5)^6} \approx - 0.984375$$

and we get for the first approximation

$${p_1} = {p_0} - {{g({p_0})} \over {g'({p_0})}} = 0.5 - {{0.039844} \over { - 0.984375}} \approx 0.540476$$

The same procedure for the second approximation give us

$${p_2} \approx 0.540476 - {{0.006085} \over { - 0.693111}} \approx 0.549 \approx 0.55$$

We shall stop at this point and approximately says that $p \approx 0.55$ , this is the answer.