4. 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. $$ 1 / 5 $$ B. $$ 5 / 24 $$ C. $$ 3 / 10 $$ D. $$ 1 / 40 $$

Question

UpStudy AI · Accepted Answer

Let the total number of components be $$N = 10$$, with $$D = 4$$ defective and $$N - D = 6$$ 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.**

$$
\text{Total ways} = \binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120
$$

**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:
  
$$
\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6
$$

- Number of ways to choose 1 non-defective component from 6:

$$
\binom{6}{1} = 6
$$

Thus, the total number of favorable outcomes is:

$$
6 \times 6 = 36
$$

**Step 3: Calculate the probability.**

$$
\text{Probability} = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{36}{120} = \frac{3}{10}
$$

The answer is $$ \frac{3}{10} $$.
The probability that two of the selected components are defective is $$ \frac{3}{10} $$, which corresponds to choice C.

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.

Upstudy AI Solution

Answer

Solution

Mind Expander

The answer is B: After all checks and balances through counts comparison.

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