Suppose that the probability of a DVD player that was manufactured in a certain factory being defective is 2%. What is the probability that 3 DVD players are defective in a shipment of 13 DVD players from this factory?

0.0019

Step-by-step explanation:

In this question, we are asked to calculate the probability that out of 13 shipments of DVD from a factory, 3 is defective.

To answer this question, what we use is the Bernoulli trial format.

Firstly, let’s write out some important information that could help us solve the question.

P(D) = 2% = 2/100 = 0.02

P(N) = 1 - P(D) = 0.98

Where P(D) is the probability of being defective and P(N) is the probability of not being defective. Now we write out the Bernoulli approximation:

P(3 defective) = 13C3. * P(D)^3 * P(N)^10

P(3 defective) = 13C3 * (0.02)^3 * (0.98)^10

P(3 defective) = 286 * 0.000008 * 0.82

P(3 defective) = 0.0019

c.

step-by-step explanation:

she should of added 9^2 and 12^2. that is because it is a^2+b^2=c^2. c^2 is the hypotenuse and should always be bigger than the other two numbers. which is why she should of added 9 squared and 12 squared together instead of 15 squared.