Probability Test for Light Bulbs
How can we calculate the probabilities of light bulbs passing or failing a test?
Given that 12 light bulbs are tested with 3 failing, what is the probability that both light bulbs failed the test? What about the probability that both light bulbs passed the test? And finally, what is the probability that at least one light bulb failed the test?
Final answer:
Probability in this problem is determined using the basic probability formula: 'Number of desired outcomes / Total outcomes'. The probability that both light bulbs fail is 1/22, that both light bulbs pass is 6/11, and that at least one light bulb fails is 5/11.
Explanation:
The subject of this question is Probability, which is a branch of math used to measure the likelihood of the occurrence of an event. To find probabilities, we often use the basic formula: Probability = Number of desired outcomes / Total number of outcomes.
(a) For two randomly chosen bulbs to both be failures, we would use the above formula. There are 3 failed light bulbs out of 12. Therefore, the first bulb being a failure is 3/12 or 1/4. If one failure is removed, we are left with 2 failures out of 11 bulbs. Therefore, the second bulb being a failure is 2/11. To get the probability that both bulbs failed the test, we multiply the probability of each event: 1/4 * 2/11 = 2/44 = 1/22.
(b) To find the probability of both light bulbs passing the test, we first need to determine how many bulbs passed the test. 12 total bulbs - 3 failures = 9 successful bulbs. Therefore, the probability that the first bulb passed is 9/12 = 3/4 and the second is 8/11 (after removing one successful bulb). Therefore, the probability that both bulbs passed is 3/4 * 8/11 = 24/44 = 6/11.
(c) Finally, the probability that at least one bulb failed the test is 1 minus the probability that both bulbs passed the test, which is 1 - 6/11 = 5/11.