Suppose 60% of a large group of animals is infected with a particular disease. What is the probability that at least 2 animals are infected in a sample of size 5?
0.0870
0.3174
0.913
0.6826​


Answer :

Answer:

0.6826

Step-by-step explanation:

Probabilities are used to determine the chances of events.

The probability that at least 2 animals are infected is (c) 0.913

The proportion of the animal infected is given as:

[tex]p =60\%[/tex]

The probability is then calculated using the following binomial equation

[tex]P(x) = ^nC_xp^x(1-p)^{n-x}[/tex]

In this case,

[tex]n = 2[/tex]

To calculate the probability that at least 2 animals are infected, we start by calculating the probability that not up to 2 animals are infected.

So, we have:

[tex]P(x<2) =P(0) + P(1)[/tex]

This gives

[tex]P(x<2) = ^5C_0 \times (60\%)^0 \times (1-60\%)^{5-0}+ ^5C_1 \times (60\%)^1 \times (1-60\%)^{5-1}[/tex]

Simplify

[tex]P(x<2) = 1 \times (60\%)^0 \times (40\%)^{5}+ 5 \times (60\%) \times (40\%)^{4}[/tex]

[tex]P(x<2) = 0.08704[/tex]

Using the complement rule, we have:

[tex]P(x \ge 2) = 1 - P(x<2)[/tex]

So, we have:

[tex]P(x \ge 2) = 1 - 0.08704[/tex]

[tex]P(x \ge 2) = 0.91296[/tex]

Approximate

[tex]P(x \ge 2) = 0.913[/tex]

Hence, the probability is (c) 0.913

Read more about probabilities at:

https://brainly.com/question/251701