a box contains 40 computer disks, 5 are defective. Three disks are selected from the box without replacement and regard to order. what is the probability that
a. all three of the disks are defective
b. none of the three disks are defective
c. exactly one of the three disks is defective
d. at least one of the three disks works
teeny weeny lol
a. What do you get when you take three defective disks out of a box? A great start to a questionable art project! The probability of getting all three defective disks is calculated by taking the number of ways to select 3 defective disks out of 5 and dividing it by the total number of ways to select any 3 disks out of 40.
Probability = (number of ways to select 3 defective disks)/(total number of ways to select any 3 disks)
b. None of the three disks are defective? That's a pretty good outcome for your computer, but not so great if you're looking for disk coasters. The probability of getting none of the disks defective is calculated by taking the number of ways to select 3 non-defective disks out of 35 (since there are 35 non-defective disks left after taking out 5 defective ones) and dividing it by the total number of ways to select any 3 disks out of 40.
Probability = (number of ways to select 3 non-defective disks)/(total number of ways to select any 3 disks)
c. Exactly one defective disk? Well, that's like having a designated screw-up in your group project. The probability of getting exactly one defective disk is calculated by taking the number of ways to select 1 defective disk out of 5 and multiplying it by the number of ways to select 2 non-defective disks out of 35 (since there are 35 non-defective disks left after taking out 5 defective ones), and then dividing it by the total number of ways to select any 3 disks out of 40.
Probability = (number of ways to select 1 defective disk)*(number of ways to select 2 non-defective disks)/(total number of ways to select any 3 disks)
d. At least one working disk? Ah, the joys of having functionality! The probability of having at least one working disk is calculated by subtracting the probability of getting all three disks defective from 1.
Probability = 1 - (probability of getting all three disks defective)
To find the probabilities, we need to use combinations. The number of ways to choose r items from a set of n items is denoted as "n choose r" and can be calculated using the combination formula:
C(n, r) = n! / (r!(n-r)!)
Now, let's solve each part of the problem step-by-step:
a. Probability that all three disks are defective:
To calculate this, we need to find the number of ways to choose 3 defective disks from the 5 available defective disks, divided by the number of ways to choose any 3 disks from the total 40 disks.
Number of ways to choose 3 defective disks: C(5, 3) = 5! / (3!(5-3)!) = 10
Number of ways to choose any 3 disks: C(40, 3) = 40! / (3!(40-3)!) = 9,880
Therefore, the probability is: 10/9,880 ≈ 0.00101
b. Probability that none of the three disks are defective:
In this case, we need to find the number of ways to choose 3 non-defective disks from the remaining 35 non-defective disks, divided by the number of ways to choose any 3 disks from the total 40 disks.
Number of ways to choose 3 non-defective disks: C(35, 3) = 35! / (3!(35-3)!) = 6545
Number of ways to choose any 3 disks: C(40, 3) = 9,880
Therefore, the probability is: 6545/9,880 ≈ 0.6628
c. Probability that exactly one of the three disks is defective:
For this case, we first need to choose 1 defective disk out of the 5 defective disks, and then choose any 2 non-defective disks out of the 35 non-defective disks.
Number of ways to choose 1 defective disk: C(5, 1) = 5
Number of ways to choose 2 non-defective disks: C(35, 2) = 35! / (2!(35-2)!) = 595
Number of ways to choose any 3 disks: C(40, 3) = 9,880
Therefore, the probability is: (5 * 595) / 9,880 ≈ 0.3015
d. Probability that at least one of the three disks works:
To calculate this probability, we'll find the complement, i.e., the probability that none of the three disks work, and then subtract it from 1.
Probability that none of the three disks work = Probability that none of the three disks are defective
From part b, we know that this probability is approximately 0.6628. Therefore:
Probability that at least one of the three disks works = 1 - Probability that none of the three disks work
= 1 - 0.6628
= 0.3372
So, the probability that at least one of the three disks works is approximately 0.3372.
To find the probabilities, we need to know the total number of outcomes and the number of favorable outcomes for each event.
Step 1: Calculate the total number of outcomes.
Since we are selecting three disks without replacement, the total number of outcomes would be the number of ways to select three disks out of 40, which can be calculated using the combination formula (nCr). It can be expressed as:
Total outcomes = 40C3 = (40!)/((40-3)!*3!)
Step 2: Calculate the number of favorable outcomes for each event.
a. Probability that all three disks are defective:
Since there are 5 defective disks in the box, the number of ways to select three defective disks can be calculated as:
Favorable outcomes (all defective) = 5C3 = (5!)/((5-3)!*3!)
b. Probability that none of the three disks are defective:
To select three non-defective disks, we need to choose them from the 35 non-defective disks in the box. So, the number of favorable outcomes becomes:
Favorable outcomes (none defective) = 35C3 = (35!)/((35-3)!*3!)
c. Probability that exactly one of the three disks is defective:
We can select one defective disk from the 5 defective disks and two non-defective disks from the 35 non-defective disks. The number of favorable outcomes is:
Favorable outcomes (exactly one defective) = 5C1 * 35C2 = (5!)/((5-1)!*1!) * (35!)/((35-2)!*2!)
d. Probability that at least one of the three disks works:
The probability of this event is the complement of the probability that none of the three disks are defective. So, we can find it using:
Probability (at least one works) = 1 - Probability (none defective)
Step 3: Calculate the probabilities for each event.
Now that we have the total number of outcomes and the favorable outcomes for each event, we can calculate the probabilities.
a. Probability (all three defective) = Favorable outcomes (all defective) / Total outcomes
b. Probability (none defective) = Favorable outcomes (none defective) / Total outcomes
c. Probability (exactly one defective) = Favorable outcomes (exactly one defective) / Total outcomes
d. Probability (at least one works) = 1 - Probability (none defective)
By substituting the values obtained in Step 2 into the respective formulas and performing the calculations, we can find the answers to each part of the question.
d - defective
g -good
a) prob(ddd) = (5/40)(4/39)(3/38) = ....
b) prob(ggg) = (35/40)(34/39)(33/38) = ....
c) prob(dgg) + prob(gdg) + prob(ggd) = 3(5/40)(35/39)(34/38) = ...
d) 1 - prob(ddd) = 1 - (35/40)(34/39)(33/38)