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ICE Princess25 [194]
3 years ago
11

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.)
Mathematics
1 answer:
aleksley [76]3 years ago
7 0

Answer:

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

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<h2><u>solución</u></h2>

deja que los números sean x, and x + 2

sabemos que su suma es igual a 100, entonces

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por lo tanto, el primer número es 49

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