# Thread: [SOLVED] Simple probability problem

1. ## [SOLVED] Simple probability problem

edit: solved

2. Originally Posted by chickeneaterguy
Suppose that Jennifer is in a final round of interviews for various jobs with four different employers, after which each employer will send out one job offer to one candidate. Suppose that two other candidates are competing with her for each of the jobs, and that every candidate has an equal probability of receiving a job offer.

a) What is the probability that Jennifer will receive at least one job offer?

b) What is the minimum number of such final interviews Jennifer needs to get to guarantee that the probability she will receive at least one job offer is more than 0.9?
Hi chickeneaterguy,

(a)

The probability Jennifer receives no offer is

$\left(\frac{2}{3}\right)^4$

Hence subtract that from 1 for the probability she gets at least one offer,
or sum all the seperate ways she can get 1, 2, 3 or 4 offers.

(b)

$1-\left(\frac{2}{3}\right)^n\ >\ 0.9$

$1-0.9\ >\ \left(\frac{2}{3}\right)^n$

$log(0.1)\ >\ log\left(\frac{2}{3}\right)^n$

$log(0.1)\ >\ nlog\left(\frac{2}{3}\right)$

As log(2/3) is negative, reverse the inequality to solve for n

$\frac{log(0.1)}{log\left(\frac{2}{3}\right)}\ <\ n$

$n\ >\ 5.67\ \Rightarrow\ n\ \ge\ 6$

3. Thanks. Can I ask why you used log though?

4. Originally Posted by chickeneaterguy
Thanks. Can I ask why you used log though?
Sure,

when the value we are looking for is in the index position,
we can extract using logs, as

$log\left(x^n\right)=n(logx)$

then we can find n.

You can also try some experimental values in this case,
as we need not expect a large value of n to be the solution.

$0.1\ >\ \left(\frac{2}{3}\right)^n$

$\frac{2}{3}=0.67$

$\left(\frac{2}{3}\right)^2=0.44$

$\left(\frac{2}{3}\right)^3=0.296$

$\left(\frac{2}{3}\right)^4=0.1975$

$\left(\frac{2}{3}\right)^5=0.1316$

$\left(\frac{2}{3}\right)^6=0.08779$

5. Okay, that makes sense. Thanks again