1. ## Probability Problem

Hey folks, I've hit my very weak spot. Probability and Statistics

I have a problem here, and I REALLY want to use $binomcdf(n,p,k)$ but it's not giving me the right answer, and I've tried using $binompdf(n,p,k)$ and $P(x success)=_{n}C_{k}P^k(1-P)^{n-k}$ all to no avail. I have the correct answer to reference, but I cannot for the life of me seem to get to it, any help would be appreciated!

The problem is this:

In one town, 74% of the homes have satellite dishes. A pollster randomly sampled 50 homeowners from the town. What is the probability that at least 42 of them have satellite dishes? Round your answer to 4 decimal places.
Right off the bat I think $P_{having a dish}=0.74$ and the first step to me seems like it should be using $binomcdf(50,0.74,42)$ because of the words at least, rather than exactly.

This isn't working at all, the correct answer to the problem is $0.0684$.

Again, any help or clarification at all would be great, thanks!

2. Originally Posted by Kasper
to use $\text{binomcdf}(n,p,k)$ but it's not giving me
The problem is this: In one town, 74% of the homes have satellite dishes. A pollster randomly sampled 50 homeowners from the town. What is the probability that at least 42 of them have satellite dishes? Round your answer to 4 decimal places. !
It the at least 42 is giving you the trouble.
$1-\text{binomcdf}(50,0.74,41)$ will give you the correct answer.

3. Originally Posted by Plato
It the at least 42 is giving you the trouble.
$1-\text{binomcdf}(50,0.74,41)$ will give you the correct answer.
I thought that binomcdf corrected the at least problem just by using cdf rather than pdf, I don't understand why we are subtracting it from one? That did work perfectly though! Thanks!

4. Originally Posted by Kasper
I don't understand why we are subtracting it from one?
Sorry I should have explained.
$\text{binomcdf}(50,0.74,42)$ is the probability of at most 42.
Its complement is $1-\text{binomcdf}(50,0.74,41)$

5. Oh I see, that makes sense. So $binompdf(n,p,k)$ is used to calculate probability of EXACTLY n events occurring, while $binomcdf(n,p,k)$ is the probability of AT MOST n events occurring. So when asked for AT LEAST, I can find this by subtracting the probability of AT MOST from 1, giving me the probability of a value occurring above that value of k, right?

6. Originally Posted by Kasper
Oh I see, that makes sense. So $binompdf(n,p,k)$ is used to calculate probability of EXACTLY n events occurring, while $binomcdf(n,p,k)$ is the probability of AT MOST n events occurring. So when asked for AT LEAST, I can find this by subtracting the probability of AT MOST from 1, giving me the probability of a value occurring above that value of k, right?
Exactly.
$\text{binom{\color{blue}CDF}}(n,p,k)$
The CDF stands for cumulative distribution function.
Cumulative means the probability accumulates as we increase the number of trials from none on.

7. Originally Posted by Plato
Exactly.
$\text{binom{\color{blue}CDF}}(n,p,k)$
The CDF stands for cumulative distribution function.
Cumulative means the probability accumulates as we increase the number of trials from none on.
Awesome, that makes so much more sense than my teachers presentation of "Use this function in this area of your calculator for exacts, and this other one for at leasts." Thanks alot!