twenty light bulbs are tested three light bulbs fail the test. two light bulbs are selected at random without replacement from this collection of bulbs find the probablilty that both light bulbs failed the test. round the answer 3 decimals places
the failure rate is 15%
the probability of both bulbs being bad is ... .15 * .15
it said "without replacement" , so the prob that the second is faulty is not 15%
Bernie, what part of my solution did you not like ?
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To find the probability of both light bulbs failing the test, we need to use the concept of conditional probability.
There are initially 20 light bulbs in total, out of which 3 have failed the test. So, there are 20 - 3 = 17 light bulbs that passed the test.
When we choose the first light bulb, the probability of selecting a failed one is 3/20, as there are 3 failed bulbs out of 20 total bulbs.
After choosing the first bulb, there are now 19 bulbs remaining in the collection, out of which 2 have failed the test. So, the probability of selecting a failed bulb as the second one is 2/19.
Since the bulbs are selected without replacement, we need to multiply the probabilities of each selection to find the probability of both light bulbs failing the test:
P(both bulbs failed) = (3/20) * (2/19)
Calculating this, we have:
P(both bulbs failed) = 0.15 * 0.1053 ≈ 0.0158
Rounded to 3 decimal places, the probability that both light bulbs failed the test is approximately 0.016.
To find the probability that both light bulbs failed the test, we first need to determine the total number of light bulbs and the number of failed light bulbs in the collection.
Given that there are 20 light bulbs in total and 3 of them failed the test, there are 20 - 3 = 17 light bulbs that passed the test.
Now, we will calculate the probability of selecting two failed light bulbs without replacement.
The probability of selecting the first failed light bulb is 3/20 since there are 3 failed light bulbs out of 20 in total.
After removing one failed light bulb, there are 19 light bulbs left, and 2 failed light bulbs remaining. So, the probability of selecting the second failed light bulb is 2/19.
To find the probability of both events occurring, we multiply the probabilities of each event together:
(3/20) * (2/19) = 6/380 = 0.0158
Therefore, the probability that both light bulbs selected at random without replacement failed the test is approximately 0.016(rounding to 3 decimal places).