Suppose that 80%, percent of trees in a forest of 300 trees are infected with a virus, and we take an SRS of 5 trees to test for the virus. What is the probability that exactly 2 of the 5 trees have the virus? You may round your answer to the nearest hundredth. P(X=2)=P(X=2)=P, left parenthesis, X, equals, 2, right parenthesis, equals

0.05

Step-by-step explanation:

Given that:

Number of samples, n = 5

Proportion, p = 80% = 0.8

P(x = 2)

Using binomial distribution formula ;

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

P(x = 2) = 5C2 * 0.8^2 * (1 - 0.8)^(5-2)

P(x = 2) = 5C2 * 0.8^2 * 0.2^3

P(x = 2) = 10 * 0.64 * 0.008

P(x = 2) = 0.0512

= 0.05 ( nearest hundredth)

hope it

step-by-step explanation:

x^(1/6) · x^(1/6) · x^(1/6)   ·x^(1/6) = x^(4/6)   since we add the exponents

4/6 can be simplified to 2/3

the simplified form is x^(2/3)

step-by-step explanation:

this is not a question