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.