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Phantasy [73]
4 years ago
13

Box 1 contains 1000 lightbulbs of which 10% are defective. Box 2 contains 2000 lightbulbs of which 5% are defective. (a) Suppose

a box is given to you at random and you randomly select a lightbulb from the box. If that lightbulb is defective, what is the probability you chose Box 1? (b) Suppose now that a box is given to you at random and you randomly select two light- bulbs from the box. If both lightbulbs are defective, what is the probability that you chose from Box 1? 4 Solve the Rainbow
Mathematics
1 answer:
mojhsa [17]4 years ago
3 0

Answer:

a) There is a 66.7% chance that you were given box 1

b) There is a 80% chance that you were given box 1

Step-by-step explanation:

To find this, we need to note that there is a 1/10 chance of getting a defective bulb with box 1 and a 1/20 chance in box 2.

a) To find the answer to this, find the probability of getting a defective bulb for each box. Since there is only one bulb pulled in this example, we just use the base numbers given.

Box 1 = 1/10

Box 2 = 1/2

From this we can see that Box 1 is twice as likely that you get a defective bulb. As a result, the percentage chance would be 2/3 or 66.7%

b) For this answer, we need to square each of the probabilities in order to get the probability of getting a defective one twice.

Box 1 = 1/10^2 = 1/100

Box 2 = 1/20^2 = 1/400

As a result, Box 1 is four times more likely. This means that it would be a 4/5 chance and have a probability of 80%

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Answer:

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Step-by-step explanation:

One of the methods of solving the given equation [(2r – 3)(2r + 3)] is the FOIL method

Using the FOIL method (Used for multiplying binomials)

F stands for first

O stands for outer

I stands for inner

L stands for last

In this method, it means we will first multiply the first terms followed by multiplying the outer term, then multiplying the inner terms and finally, the last terms.

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First : (2r)(2r) = 4r²

Outer : (2r)(+3) = 6r

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Last :  (–3)(+3) =  – 9

Adding the results we get

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