A new surgery is successful 70% of the time. If the results of 7 such surgenes are randomly sampled, what is the probability that at most 5 of them are successful?

Carry your intermediate computations to at least four decimal places, and round your answer to the deamat places.

To solve this problem, we need to calculate the probability that at most 5 out of 7 surgeries are successful. This is a binomial probability problem, where we can use the binomial distribution formula.

### Binomial Distribution Formula:
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]
Where:
- \(n\) is the number of trials (surgeries) = 7
- \(k\) is the number of successes
- \(p\) is the probability of success on each trial = 0.7
- \(1-p\) is the probability of failure = 0.3

We need to find the probability that at most 5 surgeries are successful, which means we need to sum the probabilities from \(P(X = 0)\) to \(P(X = 5)\):

\[
P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)
\]

### Step-by-Step Calculations:

1. **Calculate \(P(X = 0)\):**
\[
P(X = 0) = \binom{7}{0} (0.7)^0 (0.3)^7 = 1 \times 1 \times 0.0002187 = 0.0002187
\]

2. **Calculate \(P(X = 1)\):**
\[
P(X = 1) = \binom{7}{1} (0.7)^1 (0.3)^6 = 7 \times 0.7 \times 0.000729 = 0.0035697
\]

3. **Calculate \(P(X = 2)\):**
\[
P(X = 2) = \binom{7}{2} (0.7)^2 (0.3)^5 = 21 \times 0.49 \times 0.00243 = 0.0248627
\]

4. **Calculate \(P(X = 3)\):**
\[
P(X = 3) = \binom{7}{3} (0.7)^3 (0.3)^4 = 35 \times 0.343 \times 0.0081 = 0.0972405
\]

5. **Calculate \(P(X = 4)\):**
\[
P(X = 4) = \binom{7}{4} (0.7)^4 (0.3)^3 = 35 \times 0.2401 \times 0.027 = 0.2272485
\]

6. **Calculate \(P(X = 5)\):**
\[
P(X = 5) = \binom{7}{5} (0.7)^5 (0.3)^2 = 21 \times 0.16807 \times 0.09 = 0.3176523
\]

### Sum the Probabilities:
\[
P(X \leq 5) = 0.0002187 + 0.0035697 + 0.0248627 + 0.0972405 + 0.2272485 + 0.3176523 = 0.6707924
\]

### Final Answer:
The probability that at most 5 out of 7 surgeries are successful is approximately **0.6708** when rounded to four decimal places.

Would you like more details, or do you have any questions about this problem?

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**Related Questions:**

1. What is the probability that exactly 5 out of 7 surgeries are successful?
2. How would the probability change if the success rate was 80% instead of 70%?
3. What is the probability that all 7 surgeries are successful?
4. How does the binomial distribution apply to real-world scenarios outside of surgery?
5. Can you calculate the expected number of successful surgeries in this scenario?
6. What is the standard deviation of the number of successful surgeries in this problem?
7. How would you interpret the results if the success rate was unknown?
8. What are the assumptions behind using the binomial distribution in this case?

**Tip:** In binomial distributions, the sum of all probabilities (from \(X=0\) to \(X=n\)) always equals 1. This helps ensure calculations are correct.

Learn how to calculate the probability that at most 5 out of 7 surgeries are successful using the binomial distribution formula. This problem involves a success rate of 70% per surgery and requires calculating binomial probabilities step-by-step.

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