Question

About 11% of adult Americans are left-handed. Suppose that two people are randomly selected from this

population.
a. Create a discrete probability distribution for the number of left-handed people in a sample of two randomly
selected adult Americans.
b. What is the probability that at least one person in the sample is left-handed?

Answer 1

a)
`P(X=0)=(2C0)(0.11)^0 (1-0.11)^(2-0)=0.7921`

`P(X=1)=(2C1)(0.11)^1 (1-0.11)^(2-1)=0.1958`

`P(X=2)=(2C2)(0.11)^2 (1-0.11)^(2-2)=0.0121`

And then we have our probability distribution like this:

X | 0 | 1 | 2 |

P(X) | 0.7921 | 0.1958 | 0.0121|

b)
`P(X \geq 1) = P(X=1)+P(X=2) = 0.1958+0.0121=0.2079`

Explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

`X \sim Binom(n=20, p=0.48)`

The probability mass function for the Binomial distribution is given as:

`P(X)=(nCx)(p)^x (1-p)^(n-x)`

Where (nCx) means combinatory and it's given by this formula:

`nCx=(n!)/((n-x)! x!)`

Solution for the problem

Part a

On this case since we select a sample size of n =2 we have the following values for the number of left handed X=0,1,2. We can find the probabilities for each case since we know that p=0.11.

`P(X=0)=(2C0)(0.11)^0 (1-0.11)^(2-0)=0.7921`

`P(X=1)=(2C1)(0.11)^1 (1-0.11)^(2-1)=0.1958`

`P(X=2)=(2C2)(0.11)^2 (1-0.11)^(2-2)=0.0121`

And then we have our probability distribution like this:

X | 0 | 1 | 2 |

P(X) | 0.7921 | 0.1958 | 0.0121|

Part b

For this case we want this probability:

`P(X \geq 1) = P(X=1)+P(X=2) = 0.1958+0.0121=0.2079`