Among 6304 cases of heart pacemaker​ malfunctions, 275 were found to be caused by​ firmware, which is software programmed into the device. If the firmware is tested in 3 different pacemakers randomly selected from this batch of 6304 and the entire batch is accepted if there are no​ failures, what is the probability that the firmware in the entire batch will be​ accepted? Is this procedure likely to result in the entire batch being​ accepted?
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Part 1
The probability is 
  
enter your response here. This procedure is 
▼ 
certain
likely
unlikely
 to result in the entire batch being accepted.
​(Round to three decimal places as​ needed.)

Let's break down the problem step by step.

We have:
- A total of \(6304\) pacemakers.
- \(275\) of them have a firmware malfunction.
- The procedure involves testing \(3\) pacemakers selected randomly from this batch.
- The entire batch is accepted if **none** of the selected pacemakers have a firmware malfunction.

This is a **hypergeometric probability** problem, as we are selecting without replacement. We need to calculate the probability that all \(3\) selected pacemakers do **not** have a malfunction.

### Step 1: Define variables
- Total number of pacemakers: \( N = 6304 \)
- Number of pacemakers with malfunction (failures): \( K = 275 \)
- Number of pacemakers without malfunction (successes): \( N - K = 6304 - 275 = 6029 \)
- Number of pacemakers selected: \( n = 3 \)

### Step 2: Calculate probability of no failures
The probability of no malfunctions in all 3 selected pacemakers is:

\[
P(\text{no failures}) = \frac{\text{Number of ways to select 3 non-malfunctioning pacemakers}}{\text{Number of ways to select any 3 pacemakers}}
\]

This can be calculated as:

\[
P(\text{no failures}) = \frac{\binom{6029}{3}}{\binom{6304}{3}}
\]

### Step 3: Compute the probability
We can now calculate the individual probabilities step by step. The probability of selecting a non-malfunctioning pacemaker on each draw is as follows:

- First pacemaker: \( \frac{6029}{6304} \)
- Second pacemaker: \( \frac{6028}{6303} \) (since one has already been selected)
- Third pacemaker: \( \frac{6027}{6302} \)

Thus, the probability that all 3 selected pacemakers are non-malfunctioning is:

\[
P(\text{no failures}) = \frac{6029}{6304} \times \frac{6028}{6303} \times \frac{6027}{6302}
\]

Let me calculate that for you.The probability that all 3 randomly selected pacemakers have no firmware malfunction is approximately \(0.875\).

This procedure is **likely** to result in the entire batch being accepted, given that the probability is quite high.

### Follow-up questions:
1. How would the probability change if 5 pacemakers were tested instead of 3?
2. What is the probability of rejecting the batch if only one malfunctioning pacemaker is found?
3. How would the probability of acceptance change if the number of malfunctioning pacemakers increases?
4. What are the assumptions of the hypergeometric distribution used in this problem?
5. How would the result differ if pacemakers were selected with replacement?

**Tip:** When dealing with small sample sizes from a larger population, the hypergeometric distribution is a suitable method since the selection is without replacement.

Among 6304 cases of heart pacemaker malfunctions, 275 were found to be caused by firmware, which is software programmed into the device. If the firmware is tested in 3 different pacemakers randomly selected from this batch of 6304 and the entire batch is accepted if there are no failures, what is the probability that the firmware in the entire batch will be accepted? Is this procedure likely to result in the entire batch being accepted?

Find out the probability that a batch of pacemakers will be accepted based on random testing of 3 devices. Using the hypergeometric distribution, this problem explores the likelihood of accepting a batch when 275 out of 6304 pacemakers have firmware malfunctions. Suitable for high school students studying probability and statistics.

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