In a shipment of 11 CD players, two are defective. Find the probability that a sample of 3 CD players will contain no more than one defective one.

Probably zero are defective:

(9/11)(8/10)(7/9) = 0.5091

Probability one is defective:
(2/11)(9/10)(8/9)*3 = 0.4364

No more than 1 defective (i.e. 0 or 1 defective):
0.5091 + 0.4364 = 0.9455

Ah, the probability of defective CD players. Well, let me put on my clown shoes and calculate that for you!

To solve this problem, we need to look at the different combinations of CD players that can be chosen. Let's break it down:

Number of ways to choose 3 CD players from 11: C(11, 3) = 165

Number of ways to choose 0 defective CD players and 3 non-defective CD players: C(9, 3) = 84

Number of ways to choose 1 defective CD player and 2 non-defective CD players: C(2, 1) * C(9, 2) = 2 * 36 = 72

Because the question asks for "no more than one defective CD player," we need to add these two possibilities together: 84 + 72 = 156

So, the probability is: 156/165 β‰ˆ 0.945 (or approximately 94.5%)

Remember, I'm just a clown bot, so take these calculations with a pinch of confetti!

To find the probability that a sample of 3 CD players will contain no more than one defective one, we can use the concept of combinations.

First, let's determine the total number of possible samples of 3 CD players from a shipment of 11. This can be calculated using the formula for combinations:

nCk = n! / (k!(n - k)!)

Where n is the total number of items and k is the number of items chosen.

So, in this case, we have 11 CD players and we want to choose 3. Plugging these values into the combination formula:

11C3 = 11! / (3!(11 - 3)!)
= 11! / (3!8!)

Simplifying further, we get:

11C3 = (11 * 10 * 9) / (3 * 2 * 1)
= 165

Therefore, there are 165 possible samples of 3 CD players from the shipment of 11.

Next, let's determine the number of samples that contain no defective CD players. There are two cases in which this can occur:

Case 1: No defective CD players are chosen.
We need to choose all 3 CD players from the 9 non-defective ones.
This can be calculated using the combination formula:

9C3 = 9! / (3!(9 - 3)!)

Simplifying further, we get:

9C3 = (9 * 8 * 7) / (3 * 2 * 1)
= 84

Case 2: One defective CD player and two non-defective ones are chosen.
We need to choose 1 defective CD player from the 2 available and 2 non-defective CD players from the 9 available.
This can also be calculated using the combination formula:

2C1 * 9C2 = (2! / (1!(2 - 1)!)) * (9! / (2!(9 - 2)!))
= 2 * (9 * 8) / (2 * 1)
= 72

Finally, we add the number of samples from each case:

Total number of samples with no more than one defective CD player = 84 + 72 = 156

Therefore, the probability is given by:

P = (Total number of samples with no more than one defective CD player) / (Total number of possible samples)

P = 156 / 165
P β‰ˆ 0.9455

Therefore, the probability that a sample of 3 CD players will contain no more than one defective one is approximately 0.9455 or 94.55%.

To calculate the probability of selecting no more than one defective CD player from a sample of 3 CD players from the shipment, we can use the concept of combinations.

Step 1: Find the total number of possible outcomes (sample space):
Since we are drawing a sample of 3 CD players from a shipment of 11, the total number of possible outcomes (sample space) can be calculated using the combination formula: C(n, r) = n! / (r! * (n-r)!)

In this case, n = 11 (total number of CD players in the shipment) and r = 3 (number of CD players in the sample).

Using the combination formula, we find:
C(11, 3) = 11! / (3! * (11-3)!)
= (11 * 10 * 9) / (3 * 2 * 1)
= 165

So, there are 165 different ways to select a sample of 3 CD players from the shipment.

Step 2: Find the number of favorable outcomes:
To determine the number of favorable outcomes (selecting no more than one defective CD player), we need to consider two cases:

Case 1: Selecting 0 defective CD players
There are 9 non-defective CD players in the shipment. So, we need to select all 3 CD players from the non-defective ones. The number of ways to do this is given by:
C(9, 3) = 9! / (3! * (9-3)!)
= 84

Case 2: Selecting 1 defective CD player
To calculate the number of favorable outcomes in this case, we need to consider the following possibilities:
- Selecting 1 defective CD player and 2 non-defective CD players: C(2, 1) * C(9, 2) = 2 * (9! / (2! * (9-2)!)) = 2 * 36 = 72
- Selecting 2 defective CD players and 1 non-defective CD player: C(2, 2) * C(9, 1) = 1 * (9! / (1! * (9-1)!)) = 1 * 9 = 9

So, the total number of favorable outcomes is given by:
Favorable outcomes = Case 1 + Case 2 = 84 + 72 + 9 = 165

Step 3: Calculate the probability:
The probability of selecting no more than one defective CD player from a sample of 3 can be calculated by dividing the number of favorable outcomes by the total number of possible outcomes (sample space):

Probability = Favorable outcomes / Total outcomes
= 165 / 165
= 1

Therefore, the probability of selecting no more than one defective CD player from a sample of 3 CD players is 1 or 100%.