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Suppose we have two boxes; box 1 contains 4 defective and 16 non defective light bulbs. box 2 contains 1 defective and 1 non defective light bulb. we roll a fair die one time. if we get a 1 or a 2, we select a bulb at random from box 1. otherwise we select a bulb from box 2. what is the probability that the selected bulb will be defective?

User Josh Crews
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2 Answers

3 votes

Final answer:

The probability of selecting a defective bulb is 2/5.

Step-by-step explanation:

To find the probability that the selected bulb will be defective, we need to consider the probabilities of selecting from each box and the probability of selecting a defective bulb from each box.

First, let's calculate the probability of selecting from box 1. The probability of rolling a 1 or a 2 on a fair die is 2/6 or 1/3. Therefore, the probability of selecting from box 1 is 1/3.

The probability of selecting a defective bulb from box 1 is 4/20 or 1/5.

Next, let's calculate the probability of selecting from box 2. The probability of rolling anything other than a 1 or a 2 is 4/6 or 2/3. Therefore, the probability of selecting from box 2 is 2/3.

The probability of selecting a defective bulb from box 2 is 1/2.

Now, we can calculate the overall probability of selecting a defective bulb. We need to sum the probabilities of selecting from each box multiplied by the probabilities of selecting a defective bulb from each box:

(1/3) * (1/5) + (2/3) * (1/2) = 1/15 + 2/6 = 1/15 + 1/3 = 1/15 + 5/15 = 6/15 = 2/5

User ChrKoenig
by
8.5k points
6 votes
1/24 chance
Your welcome
User Mahdi Rashidi
by
7.9k points
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