A quality control inspector is to inspect 15 tools 2 of which are 
defective by selecting 4 at random.
a) What is the probability that there will be one defective tool among 
the 4 randomly selected?
b) What is the probability that there will be no defective tools among 
the 4 randomly selected?
c) How many non-defective tools do we need to add, to the tools to 
inspect, in order that the probability of having non defective tools 
among the 4 selected is 0.70?

This problem involves probability with combinations, where we are randomly selecting tools and calculating probabilities based on different scenarios. Let's tackle the problem step by step:

### Given:
- Total tools: 15
- Defective tools: 2
- Non-defective tools: \( 15 - 2 = 13 \)
- Tools selected: 4

#### Part (a): Probability of selecting **one defective tool** among the 4 selected

We need to find the probability of selecting exactly **1 defective tool** and **3 non-defective tools** from the 15 tools.

The total number of ways to select 4 tools from 15 is given by the combination formula:
\[
\text{Total ways to select 4 tools} = \binom{15}{4}
\]

The number of ways to select 1 defective tool from 2 defective tools is:
\[
\text{Ways to select 1 defective} = \binom{2}{1}
\]

The number of ways to select 3 non-defective tools from 13 non-defective tools is:
\[
\text{Ways to select 3 non-defective} = \binom{13}{3}
\]

Thus, the probability is:
\[
P(\text{1 defective}) = \frac{\binom{2}{1} \cdot \binom{13}{3}}{\binom{15}{4}}
\]

#### Part (b): Probability of selecting **no defective tools** among the 4 selected

We need to find the probability of selecting all 4 tools as **non-defective**.

The number of ways to select 4 non-defective tools from 13 non-defective tools is:
\[
\text{Ways to select 4 non-defective} = \binom{13}{4}
\]

Thus, the probability is:
\[
P(\text{no defective}) = \frac{\binom{13}{4}}{\binom{15}{4}}
\]

#### Part (c): How many non-defective tools need to be added to achieve a 0.70 probability of selecting only non-defective tools?

Let \( x \) be the number of additional non-defective tools added to the original 15 tools. The new total number of tools becomes \( 15 + x \), and the number of non-defective tools becomes \( 13 + x \).

The probability of selecting 4 non-defective tools from this new set must equal 0.70:
\[
P(\text{no defective}) = \frac{\binom{13+x}{4}}{\binom{15+x}{4}} = 0.70
\]

We will solve for \( x \).

Let me calculate these results.### Results for parts (a) and (b):
- **(a)** The probability of selecting exactly **one defective tool** among the 4 selected is approximately **0.419**.
- **(b)** The probability of selecting **no defective tools** among the 4 selected is approximately **0.524**.

Now, for part (c), let's calculate how many additional non-defective tools \(x\) need to be added to achieve a probability of 0.70 for selecting no defective tools.

We need to solve the equation:
\[
\frac{\binom{13+x}{4}}{\binom{15+x}{4}} = 0.70
\]
I'll calculate that next.### Part (c) Result:
You would need to add approximately **11 non-defective tools** to achieve a 0.70 probability of selecting no defective tools among the 4 randomly chosen.

Would you like more details on any part of the calculations or explanations?

Here are 5 related questions:
1. What is the probability of selecting exactly 2 defective tools?
2. How would the probabilities change if we inspected 5 tools instead of 4?
3. What if the number of defective tools was 3 instead of 2?
4. How would the probability calculations change if tools were inspected without replacement?
5. Can you apply a similar approach for different sample sizes?

**Tip:** When dealing with combinations, remember that the order of selection does not matter, unlike permutations where order does.

A quality control inspector is to inspect 15 tools 2 of which are defective by selecting 4 at random. a) What is the probability that there will be one defective tool among the 4 randomly selected? b) What is the probability that there will be no defective tools among the 4 randomly selected? c) How many non-defective tools do we need to add, to the tools to inspect, in order that the probability of having non defective tools among the 4 selected is 0.70?

This problem involves calculating probabilities in a quality control scenario where 15 tools are inspected, 2 of which are defective. It covers finding the probability of selecting one defective tool, no defective tools, and determining the number of non-defective tools required for a 0.70 probability. Suitable for students in Grades 10-12.

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