Question
- A box contains 10 identical electronic components of which 4 are defective. If 3 components are selected at random from the box in succession without replacing the units already drawn, what is the probability that two of the selected components are defective?
A.
B.
C.
D.
A.
B.
C.
D.
Ask by Ortega Garza. in South Africa
Mar 12,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The probability that two of the selected components are defective is
, which corresponds to choice C.
Solution
Let the total number of components be
, with
defective and
non-defective components. We select 3 components without replacement. We need the probability that exactly 2 of the selected components are defective.
Step 1: Count the total number of ways to choose 3 components from 10.
Step 2: Count the favorable outcomes.
We need to choose exactly 2 defective components and 1 non-defective component.
- Number of ways to choose 2 defective components from 4:
- Number of ways to choose 1 non-defective component from 6:
Thus, the total number of favorable outcomes is:
Step 3: Calculate the probability.
The answer is
.
Answered by UpStudy AI and reviewed by a Professional Tutor
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Mind Expander
To find the probability that two out of the three selected components are defective, we can use combinations.
First, we find the total number of ways to choose 3 components from 10. This is given by:
Next, we calculate the number of ways to choose 2 defective components out of the 4 defective ones and 1 non-defective from the remaining 6. This is:
Now, the probability of selecting 2 defective and 1 non-defective component is:
However, this is incorrect as we intended just two defective in three total. The correct approach should just focus on correctly counting these events.
Instead, using the scenarios,
- Choose 2 defective from 4:
- Choose 1 non-defective from 6:
Multiplying these gives us the favorable outcomes divided by total outcomes.
Thus, the correct answer matches scenario realistic checks:
From the earlier calculations:
Calculating all probabilities indicates
, showing the needed bias of details.
Calculating all probabilities indicates
Hence, the ultimate probability asked for landing checks reveals:
The probability that exactly 2 of the selected components are defective is
B.
B.
Thus: