# Math Help - Probability of Students being left handed..

1. ## Probability of Students being left handed..

I just took my stat final exam and flunked it, so math will be my new hobby until i feel comfortable with it, here is one of the problems that stumped me less than 30 minutes ago:

10% of the population is left handed, out of 7 students what is:

The Probability that at least one of the students is left handed?

And

The probability that at most one of the students is left handed?

At fist i assumed i would just do .10^7...but then i was not sure if i should be using the binomial distribution or not and that threw me way off...

Expect to see me a lot more around here, i'm hungry for knowledge.

Thanks again guys/gals,
Frank

2. Originally Posted by Born4spd
...but then i was not sure if i should be using the binomial distribution or not and that threw me way off...

Hi Frank,

This was a good idea. X is Bi(7,0.1) so find $P(X\leq 1) = P(X=0)+P(x=1)$

where $P(X=k) = ^nC_k\times p^k \times (1-p)^{n-k}$

Hope to see you again,

3. Im sorry i guess i should have been more specific as to where i got lost in the process of using the binomial distribution:
$7_C_k\times .10^k \times (1-.10)^{7-k}$

First time using the coding i hope it comes out right, i could not figure out what K was while i was in class.

I tried plugging in all of the numbers 1-7 for k but it kept coming out really small so i assumed it was incorrect.
Here's what i did:I was told that when it says "at least" we had to calculate for everything equal to and above that number and i remembered that when ever you want to find the probability of something "or" another happening you simply add the together...?

$_ 7C_1\times .10^1 \times (1-.10)^{7-1} =.3720087$
$_ 7C_2\times .10^2 \times (1-.10)^{7-2} =.1240029$
$_ 7C_3\times .10^3 \times (1-.10)^{7-3} =.0229635$
$_ 7C_4\times .10^4 \times (1-.10)^{7-4} =.0025515$
$_ 7C_5\times .10^5 \times (1-.10)^{7-5} =1.70E-4$
$_ 7C_6\times .10^6 \times (1-.10)^{7-6} =6.3E-6$
$_ 7C_7\times .10^7 \times (1-.10)^{7-7} =1E-7$

and i was going to add the answers to each of those together to get the probability of any of them occurring:and that gave me $.521703$

So how far off am i?

It doesn't seem right at all that there is a $52%$ chance that atleast one of the students is left handed?

4. For 'at least'

$\displaystyle P(X\geq 1) = 1-P(X=0) = 1 - ^7C_0\times 0.1^0 \times (1-0.1)^{7-0}$

Post #2 describes 'at most'

5. ## left handers probability

Originally Posted by Born4spd
I just took my stat final exam and flunked it, so math will be my new hobby until i feel comfortable with it, here is one of the problems that stumped me less than 30 minutes ago:

10% of the population is left handed, out of 7 students what is:

The Probability that at least one of the students is left handed?

And

The probability that at most one of the students is left handed?

At fist i assumed i would just do .10^7...but then i was not sure if i should be using the binomial distribution or not and that threw me way off...

Expect to see me a lot more around here, i'm hungry for knowledge.

Thanks again guys/gals,
Frank
In my opinion, the probability is: p=1-[90C7]/[100C7] which means:
1 - [90!/(7!*83!)]/[100!/(7!*93!)]. I tried to calculate it and I found:
p = 0.53325. The probability is 53.325% of having at least one left hander out of 7 students.

friendly,

Nicos Mavrommatis,
School - teacher, Haliartos Greece

6. Originally Posted by Nicromm4

In my opinion, the probability is: p=1-[90C7]/[100C7] which means:
1 - [90!/(7!*83!)]/[100!/(7!*93!)]. I tried to calculate it and I found:
p = 0.53325. The probability is 53.325% of having at least one left hander out of 7 students.

friendly,

Nicos Mavrommatis,
School - teacher, Haliartos Greece

Assuming the population is 'large' (and when a question says population but does not specify a number then it's a good bet the person who wrote the question expects this assumption to be made) then the binomial distribution is the approach to use. Several posts have already explained how to do this.

Otherwise the hypergeomtric would have to be used in some way ....

7. Good evening,

Assume you've got 100 balls in the urn. 90 green (right handers), 10 yellow (lefties). You pick simiultaneously at random 7 balls. What is the probability p of having at least 1 yellow?
Well, p = 1 - q (=the probability of having all 7 green balls). How many groups of 7 green balls are there? 7 out of 90. Which means: C'=90!/(7!*83!). How many groups of 7 balls can you combine? 7 out of 100: C=100!/(7!*93!). So, q = C'/C, green colour all seven. The rest (1-q) have at least one yellow.
OK, I considered that left handers are 10% of a population.

Thank you again,

Nicos

8. Originally Posted by Nicromm4
Good evening,

Assume you've got 100 balls in the urn. 90 green (right handers), 10 yellow (lefties). You pick simiultaneously at random 7 balls. What is the probability p of having at least 1 yellow?
Well, p = 1 - q (=the probability of having all 7 green balls). How many groups of 7 green balls are there? 7 out of 90. Which means: C'=90!/(7!*83!). How many groups of 7 balls can you combine? 7 out of 100: C=100!/(7!*93!). So, q = C'/C, green colour all seven. The rest (1-q) have at least one yellow.
OK, I considered that left handers are 10% of a population.

Thank you again,

Nicos
So you're assuming a specific population size and then using a hypergeometric distribution? As has been remarked several times in this thread, I think the question will expect a student to assume a population large enough that the binomial disribution can be used.