Atest for a disease gives an answer "positive" or "negative". because of experimental error, the test gives an incorrect answer with probability 1/4. that is, if the person has the disease, the outcome is negative with probability 1/4 and if the person does not have the disease, the outcome is positive with probability 1/4. you can assume that the probability an error is made in one test is independent of the outcomes of any other tests. in order to improve the probability of getting the correct answer, the test is run n times and the majority outcome is used as the final answer. if n = 3, what is the probability of getting the wrong answer? what about n = 5?

1. n = 3, P(x >= 2) = 0.15625  2. n = 5, P(x >= 3) = 0.103491

Step-by-step explanation:

Consider that P = ¼, by that logic Q = 1 – P = 1 – ¼ = ¾, where Q represents the value of the probability of the result being correct

When n = 3, the wrong probability can be written as p(x ≥ 2)

P(x ≥ 2) = P(x = 2) + P(x = 3)

P(x ≥ 2) = ³C₂ (1/4)² x (3/4) + (1/4)³

P(x ≥ 2) = 0.140625 + 0.015625

P(x ≥ 2) = 0.15625

When n = 5 the wrong probability P(x >= 3)

P(x ≥ 3) = P(x = 3) + P(x = 4) + P(x = 5)

P(x ≥ 3) = ⁵C₃ x (1/4)³ x (3/4)² + ⁵C₄ (1/4)⁴ x  (3/4) + (1/4)⁵

P(x ≥ 3) = (10 x 0.15625 x 0.5625) + (5 x 0.00390 x 0.75) + (0.000976)

P(x ≥ 3) = 0.08789 + 0.014625 + 0.000976

P(x ≥ 3) = 0.103491

on a coordinate grid, point a is at (−4.0, −2.4) and point b is at (4.0, −2.4). point b is a reflection of point a across the –axis. input either a lowercase x or i think is the right awnser

step-by-step explanation: