a. There are a total of 12 light bulbs in the box, with 4 being defective and 8 being non-defective. The possible outcomes when you draw 3 bulbs would be:
DDD, DDN, DND, NDD, DNN, NDN, NND, NNN (where D is a defective bulb and N is a non-defective bulb).
b. The probability that all 3 bulbs chosen are defective is:
(4/12) * (3/11) * (2/10) = 1/55, or approximately 0.0182 (1.82%).
c. There are 3 different ways to choose exactly 2 defective bulbs and 1 non-defective bulb:
- You could choose a defective bulb first, then another defective bulb second, then a non-defective bulb third (i.e., DDN).
- You could choose a defective bulb first, then a non-defective bulb second, then another defective bulb third (i.e., DND).
- You could choose a non-defective bulb first, then a defective bulb second, then another defective bulb third (i.e., NDD).
The probability of each of these possibilities is:
- DDN: (4/12) * (3/11) * (8/10) = 8/165
- DND: (4/12) * (8/11) * (3/10) = 8/165
- NDD: (8/12) * (4/11) * (3/10) = 12/165
These are all mutually exclusive events, so to find the probability of exactly two of the three bulbs being defective, we add these probabilities together:
8/165 + 8/165 + 12/165 = 28/165, or approximately 0.1697 (16.97%).
d. To find the probability that at least one of the three bulbs is defective, you can find the complement of the event "all three bulbs are non-defective" (i.e., NNN).
The probability of choosing all three non-defective bulbs is:
(8/12) * (7/11) * (6/10) = 14/55
So the probability of at least one bulb being defective is:
1 - 14/55 = 41/55, or approximately 0.7455 (74.55%).
A box contains 12 lights bulbs, of which 4 are defective. All bulbs look alike and have equal chance of being chosen. 3 light bulbs are selected from the box to check the quality.
a. List all the possible outcomes
b. find the probability that all 3 are defective
c. Find the probability that exactly two of the three chhosen are defective
d. Find the probability that a least one of the three are defective.
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