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.

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. To list all the possible outcomes, we need to consider all the different ways we can choose 3 light bulbs from the box. Since all bulbs look alike, each selection will be considered unique regardless of the order.

Possible outcomes (ignoring order):
1. All 3 bulbs are defective
2. 2 bulbs are defective and 1 is not defective
3. 1 bulb is defective and 2 are not defective
4. None of the bulbs are defective

b. To find the probability that all 3 bulbs are defective, we need to find the ratio of the number of outcomes where all 3 selected bulbs are defective to the total number of possible outcomes.

Number of outcomes where all 3 are defective: 4C3 = 4 (since there are 4 defective bulbs to choose from and we want to choose all 3)
Total number of possible outcomes: 12C3 = (12*11*10)/(3*2*1) = 220

Probability = Number of favorable outcomes / Total number of possible outcomes
= 4/220
= 1/55

c. To find the probability that exactly two of the three chosen bulbs are defective, we need to find the ratio of the number of outcomes where 2 bulbs are defective and 1 is not defective to the total number of possible outcomes.

Number of outcomes where exactly 2 bulbs are defective: 4C2 * 8C1 = (4*3)/(2*1) * 8 = 48
Total number of possible outcomes: 12C3 = 220

Probability = Number of favorable outcomes / Total number of possible outcomes
= 48/220
= 6/55

d. To find the probability that at least one of the three chosen bulbs is defective, we need to find the ratio of the number of outcomes where at least one bulb is defective to the total number of possible outcomes.

Number of outcomes where at least 1 bulb is defective = Total number of outcomes - Number of outcomes where no bulb is defective

Number of outcomes where no bulb is defective: 8C3 = (8*7*6)/(3*2*1) = 56
Total number of possible outcomes: 12C3 = 220

Number of outcomes where at least 1 bulb is defective = 220 - 56 = 164

Probability = Number of favorable outcomes / Total number of possible outcomes
= 164/220
= 41/55

To answer these questions, we need to understand the concept of combinations and probabilities.

In this problem, we have a box with 12 light bulbs, 4 of which are defective. We will select 3 light bulbs at random to check their quality. Since all the bulbs look alike and have an equal chance of being chosen, we can use combinations to find the number of possible outcomes.

a. List all the possible outcomes:
To find all the possible outcomes, we need to determine the number of combinations of 3 bulbs that can be chosen from a set of 12 bulbs.

The number of combinations can be calculated using the formula:
C(n, r) = n! / (r! * (n-r)!)

Where n is the total number of items, and r is the number of items being chosen. In our case, n = 12 (total number of bulbs) and r = 3 (number of bulbs being selected).

Using the formula, we can calculate the number of combinations:
C(12, 3) = 12! / (3! * (12-3)!) = 220

Therefore, there are 220 possible outcomes when selecting 3 bulbs from the box.

b. Find the probability that all 3 bulbs are defective:
Since there are 4 defective bulbs in the box, the probability of selecting a defective bulb on the first draw is 4/12 (4 defective bulbs out of 12 total bulbs).

Once a bulb is chosen, there are now 11 remaining bulbs in the box, with 3 of them being defective. Therefore, the probability of selecting a defective bulb on the second draw is 3/11.

Similarly, for the third draw, there are now 10 remaining bulbs in the box, with 2 being defective. So, the probability of selecting a defective bulb on the third draw is 2/10.

To find the probability that all 3 bulbs are defective, we need to multiply the individual probabilities together:
P(all 3 bulbs defective) = (4/12) * (3/11) * (2/10) = 1/110

Therefore, the probability that all 3 bulbs selected are defective is 1/110.

c. Find the probability that exactly two of the three selected bulbs are defective:
To calculate this probability, we need to consider the different scenarios where exactly two bulbs out of the three are defective.

Scenario 1: The first bulb selected is defective, the second bulb selected is also defective, and the third bulb selected is non-defective.
The probability of this scenario is given by:
P(defective, defective, non-defective) = (4/12) * (3/11) * (8/10) = 24/1320

Scenario 2: The first bulb selected is non-defective, the second bulb selected is defective, and the third bulb selected is also defective.
The probability of this scenario is given by:
P(non-defective, defective, defective) = (8/12) * (4/11) * (3/10) = 96/1320

Scenario 3: The first bulb selected is defective, the second bulb selected is non-defective, and the third bulb selected is also defective.
The probability of this scenario is given by:
P(defective, non-defective, defective) = (4/12) * (8/11) * (3/10) = 96/1320

Adding up the probabilities of all these scenarios:
P(exactly two bulbs defective) = P(defective, defective, non-defective) + P(non-defective, defective, defective) + P(defective, non-defective, defective)
= (24/1320) + (96/1320) + (96/1320)
= 216/1320

Therefore, the probability that exactly two of the three selected bulbs are defective is 216/1320.

d. Find the probability that at least one of the three selected bulbs is defective:
To find the probability that at least one bulb is defective, we need to consider the opposite scenario - i.e., the probability that none of the bulbs are defective and subtract it from 1.

The probability of selecting a non-defective bulb on the first draw is 8/12.
On the second draw, with one non-defective bulb having been removed from the box, the probability of selecting another non-defective bulb is 7/11.
Similarly, on the third draw, with two non-defective bulbs having been removed, the probability of selecting a non-defective bulb is 6/10.

To find the probability of selecting three non-defective bulbs, we multiply the individual probabilities together:
P(no defective bulbs) = (8/12) * (7/11) * (6/10) = 336/1320

Therefore, the probability that at least one of the three selected bulbs is defective is:
P(at least one defective bulb) = 1 - P(no defective bulbs)
= 1 - (336/1320)
= 984/1320

Therefore, the probability that at least one of the three selected bulbs is defective is 984/1320.