A firm manufactures light bulbs and those that come out of the assembly line are either defective or good. Every five bulbs that come out, 1 is defective. If 3 bulbs are selected at random, what is the probability that:
(i) All 3 are defective
(ii) All 3 are good
(iii) The first selected is defective
(iv) 1 defective and 2 good
(v) 2 defective and 1 good
(vi) At most 2 are good
If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
1. (1/5)^3 = ?
2. (4/5)^3 = ?
3. 1/5
4. 1/5 * 4/5 * 4/5 = ?
5. You do it, using similar process.
6. This is none good, one good or 2 good.
Either-or probabilities are found by adding the individual probabilities.
To answer these probability questions, we need to consider the total number of outcomes and the favorable outcomes.
First, let's find the total number of outcomes. We need to determine the total number of ways to select 3 bulbs from the available ones. Since each bulb can either be defective or good, we have 2 options for each bulb. Hence, the total number of outcomes can be calculated using the formula for combinations:
Total outcomes = C(n, r) = n! / (r!(n-r)!),
where n is the total number of bulbs and r is the number of bulbs selected.
In this case, since we have 5 bulbs coming out of the assembly line and we are selecting 3 bulbs, the total number of outcomes will be:
Total outcomes = C(5, 3) = 5! / (3!(5-3)!) = 10.
Now, let's calculate the favorable outcomes for each probability question:
(i) All 3 are defective: In this case, we need to select all 3 defective bulbs. Since 1 out of every 5 bulbs is defective, the probability of selecting a defective bulb is 1/5. Since we need 3 defective bulbs in a row, we multiply these probabilities together:
Favorable outcomes = (1/5) * (1/5) * (1/5) = 1/125.
(ii) All 3 are good: In this case, we need to select all 3 good bulbs. The probability of selecting a good bulb is 4/5, as only 1 out of every 5 bulbs is defective. Multiplying these probabilities together, we get:
Favorable outcomes = (4/5) * (4/5) * (4/5) = 64/125.
(iii) The first selected is defective: In this case, the first bulb needs to be defective and the other two can be any bulbs (defective or good). The probability of selecting a defective bulb is 1/5, and for the next two bulbs, it doesn't matter if they are defective or good. Hence,
Favorable outcomes = (1/5) * 1 * 1 = 1/5.
(iv) 1 defective and 2 good: In this case, we need to select 1 defective bulb and 2 good bulbs. The probabilities can be calculated as follows:
Favorable outcomes = (1/5) * (4/5) * (4/5)
+ (4/5) * (1/5) * (4/5)
+ (4/5) * (4/5) * (1/5)
= 48/125.
(v) 2 defective and 1 good: In this case, we need to select 2 defective bulbs and 1 good bulb. The probabilities can be calculated as follows:
Favorable outcomes = (1/5) * (1/5) * (4/5)
+ (1/5) * (4/5) * (1/5)
+ (4/5) * (1/5) * (1/5)
= 24/125.
(vi) At most 2 are good: This means either all 3 bulbs are defective or only 1 bulb is good and the other two are either defective or good. We can add the probabilities of these two cases:
Favorable outcomes = 1/125 + 48/125 = 49/125.
Therefore, the probabilities are:
(i) All 3 are defective: 1/125
(ii) All 3 are good: 64/125
(iii) The first selected is defective: 1/5
(iv) 1 defective and 2 good: 48/125
(v) 2 defective and 1 good: 24/125
(vi) At most 2 are good: 49/125.