A shipment of 20 DVDs has arrived at a video rental store. Based on past experience, the manager knows that 10% of all new DVDs sent to the store have a visible defect. The manager tells you to begin inspecting the new DVDs one at a time at random until you find the first DVD that has a defect. If 10% of the DVDs have a visible defect in the new shipment, what is the probability that the first DVD that has a defect is the 3rd one that you inspect? (Round your answer to 3 decimal places.)

The probability is 0.081

Step-by-step explanation:

Here, we want to calculate the probability that the 3rd inspection will be defective.

What this means is that the first two would

not be defective.

Probability of having a defective DVD = 10% = 10/100 = 0.1

Probability of not having a defective DVD = 1 -0.1 = 0.9

So the probability of third being defective = Probability of first not defective * Probability of second not defective * Probability of third defective

= 0.9 * 0.9 * 0.1 = 0.081

15

step-by-step explanation:

we are given the following expression and we are to solve it and find the product:

-13 × (-5)

solving the given expression in the standard order of solving a mathematical expression, beginning with the opening of brackets and then multiplying the terms and changing their signs.

-3 × -5 = 15 (answer is positive because when negative is multiplied with negative, it gives a positive answer)