# help with probability

• Jul 6th 2008, 12:51 AM
hereami
help with probability
Hello!
I have two questions in the area of probability related to a development course being taken, but I never took this stuff back in college and don't know where to begin! Any help would be oh so appreciated.

Here they are:
1. "Fifty people answer calls in a call centre. The pobability that any individual will be absent is 0.02. What is the probability that no more than two people will be absent on a typical day?"

-Now, I do remember that the probability of exactly two independent events A and B happening is figured by multiplying the P(A) by the P(B). But the "no more than" is what has me confused. Where do I start?

2. On average, five calls per hour will not be answered within the target time. What is the probability of:
i)six calls not being answered within the target time during one hour?
ii) four calls not being answered within the target time during one hour?
iii)a total of ten calls not being answered within the target time during a two-hour period?

-This one has me totally lost from the beginning.

Any help would be great. Thanks!
• Jul 6th 2008, 02:00 AM
Jhevon
Quote:

Originally Posted by hereami
Hello!
I have two questions in the area of probability related to a development course being taken, but I never took this stuff back in college and don't know where to begin! Any help would be oh so appreciated.

Here they are:
1. "Fifty people answer calls in a call centre. The pobability that any individual will be absent is 0.02. What is the probability that no more than two people will be absent on a typical day?"

-Now, I do remember that the probability of exactly two independent events A and B happening is figured by multiplying the P(A) by the P(B). But the "no more than" is what has me confused. Where do I start?

first note that we have independent trials here, since we are dealing on an individual basis and the absence of one person doesn't affect the others. so then, we can use Bernoulli trials (the Binomial distribution), since we have independent trials and two possible outcomes per trial that can be thought of as success or failure (namely, absent or present).

By this method, the probability of $k$ successes in $n$ trials, where the probability of success is $p$ and the probability of failure is $q = 1 - p$ is given by:

$P(k) = {n \choose k} p^kq^{n - k}$

let "absent" be our "success" here, since that is what we are looking for. then the probability of success is p = 0.02 and the probability of failure is q = 1 - 0.02 = 0.98. we have n = 50 trials.

now, we want the probability that no more than 2 are absent, that means, 2 or less would be absent. that is, the probability that 2 people are absent, OR 1 person is absent, OR no one is absent. thus you want:

$P(\text{absent} \le 2) = P(0) + P(1) + P(2) =$ ${50 \choose 0}(0.02)^0(0.98)^{50} + {50 \choose 1}(0.02)^1(0.98)^{49} + {50 \choose 2}(0.02)^2(0.98)^{48} = \cdots$

i leave the computations to you :p
• Jul 6th 2008, 05:41 AM
janvdl
Quote:

Originally Posted by hereami
2. On average, five calls per hour will not be answered within the target time. What is the probability of:
i)six calls not being answered within the target time during one hour?
ii) four calls not being answered within the target time during one hour?
iii)a total of ten calls not being answered within the target time during a two-hour period?

If they speak of averages, shift your mind to the Poisson distribution.

So we have:
$\lambda = 5$

i) $k = 6$

So:

$P(6) = \frac{\lambda ^{k} e^{- \lambda}}{k!} = \frac{5 ^{6} e^{- 5}}{6!}$

============================

ii) $k = 6$

So:

$P(4) = \frac{\lambda ^{k} e^{- \lambda}}{k!} = \frac{5 ^{4} e^{- 5}}{4!}$

============================

iii) The trick here is that the time has changed. It is now 2 hours.
Originally we had an average of 5 per hour. So we will have 10 per 2 hours.
Lambda is now 10.

$P(10) = \frac{\lambda ^{k} e^{- \lambda}}{k!} = \frac{10 ^{10} e^{- 10}}{10!}$