# Thread: Probabilty having do with left-handed people

1. ## Probabilty having do with left-handed people

Gia learned that about 10% of people are left‐handed. She ran 10 different simulations using random digits to find the probability that there is a left-handed person in a group of 5 randomly selected people.

In the table below, 0 represents a left‐handed person and 1 through 9 represent a right‐handed
person. Each row represents one simulation of 5 people.
3 1 4 9 6
5 6 9 6 7
5 3 2 6 8
1 4 4 2 8
9 4 4 2 6
6 8 7 4 9
6 3 4 8 8
2 7 4 4 2
0 3 6 1 6
0 5 2 0 9
Based on Gia's simulations, what is the probability that in a group of 5 people, at least 1 person will be left‐handed?

2. ## Re: Probabilty having do with left-handed people

I would consider the 10 simulations as a single one and estimate the probability of a lefty from that.

In this case it appears to be $p = \dfrac{3}{50}$

Then $P[\text{at least 1 lefty in 5 people}] = 1 - P[\text{no lefties in 5 people}] = 1 - (1-p)^5 = \dfrac{83154993}{312500000} \approx 0.266$

3. ## Re: Probabilty having do with left-handed people

Originally Posted by romsek
I would consider the 10 simulations as a single one and estimate the probability of a lefty from that.

In this case it appears to be $p = \dfrac{3}{50}$

Then $P[\text{at least 1 lefty in 5 people}] = 1 - P[\text{no lefties in 5 people}] = 1 - (1-p)^5 = \dfrac{83154993}{312500000} \approx 0.266$
I can see what you are doing but don't think that answers the question.

The "10% of people are lefthanders" is built into the simulation (0=left, 1 to 9 = right).

Based on the simulations (where 2 out of 10 groups of 5 have at least 1 lefthander) the point estimate of the probability required would be 2/10 = 0.2.

4. ## Re: Probabilty having do with left-handed people

Originally Posted by Debsta
I can see what you are doing but don't think that answers the question.

The "10% of people are lefthanders" is built into the simulation (0=left, 1 to 9 = right).

Based on the simulations (where 2 out of 10 groups of 5 have at least 1 lefthander) the point estimate of the probability required would be 2/10 = 0.2.
Good point. But if $p=0.2$ is hardwired in then we can simply apply this to obtain

$P[\text{no lefties in 5}] = 1 - (1-0.2)^5 \approx 0.67$

The whole problem smacks of the writer not really understanding probability very well.

5. ## Re: Probabilty having do with left-handed people

The whole problem smacks of the writer not really understanding probability very well.
Totally agree!!

6. ## Re: Probabilty having do with left-handed people

Thanks for answering to my question! This was on my daughter's 7th grade worksheet and the teacher claimed the answer was 6% (3/15) which was probably on the answer key. However, I had a hard time believing this was the answer. Thanks again!

7. ## Re: Probabilty having do with left-handed people

Originally Posted by vlthompso
Thanks for answering to my question! This was on my daughter's 7th grade worksheet and the teacher claimed the answer was 6% (3/15) which was probably on the answer key. However, I had a hard time believing this was the answer. Thanks again!
But 6% is not the same as 3/15 ??