Question

A recent Harvard study shows that approximately 12% of Americans are left‑handed. Suppose Alex takes a random sample of four Americans. What is the probability that at least one American is left‑handed or at least three Americans are right‑handed?

Answer 1

Answer:

A. The probability of at least one American is left-handed is 92.68% for a sample of 4 random Americans.

B. The probability of at least three Americans are right-handed is 7.32% for a sample of 4 random American

Step-by-step explanation:

This is a binomial distribution study where n = 4 and p = 0.12. Using the binomial table, where the first column is the distribution function, f(x), and the second is the cumulative distribution function F(x), we can calculate the probabilities.

0 0.5997 0.5997

1 0.3271 0.9268

2 0.0669 0.9937

3 0.0061 0.9998

4 0.0002 1.0000

Now, we can answer the questions, this way:

A. What is the probability that at least one American is left‑handed?

P (0) + P (1) = 0.5997 + 0.3271 = 0.9268

The probability of at least one American is left-handed is 92.68% for a sample of 4 random Americans.

B. What is the probability that at least three American are right‑handed?

1 - P(0) - P(1) = 1 - 0.5997 - 0.3271 = 0.0732

The probability of at least three Americans are right-handed is 7.32% for a sample of 4 random Americans.

Answer 2

Answer:

132.71% probability that at least one American is left‑handed or at least three Americans are right‑handed

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they are right handed, or they are left handed. The probability of a person being left or right handed is independent of any other person. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Sample of four:

This means that $n = 4$

At least one left-handed

12% are left handed, so p = 0.12.

Either there are no left handed people in the sample, or there is at least one. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$, so

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{4,0}.(0.12)^{0}.(0.88)^{4} = 0.5997$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.5997 = 0.4003$

At least three right-handed:

100-12 = 88% are right handed, so p = 0.88.

$P(X \geq 3) = P(X = 3) + P(X = 4)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{4,3}.(0.88)^{3}.(0.12)^{1} = 0.3271$

$P(X = 4) = C_{4,4}.(0.88)^{4}.(0.12)^{0} = 0.5997$

What is the probability that at least one American is left‑handed or at least three Americans are right‑handed?

0.4003 + 0.3271 + 0.5997 = 1.3271

132.71% probability that at least one American is left‑handed or at least three Americans are right‑handed