Define binomial distribution. During one stage in the manufacture of integrated circuit chips, a coating must be applied. If 70% of chips received a thick enough coating, find the probability that among 15 chips

(i) at least 12 will have thick enough coatings,

(ii) exactly 10 will have thick enough coatings.

### Binomial Distribution

The binomial distribution is a discrete probability distribution that models the number of successes in a fixed number of independent trials of a binary experiment. Each trial results in one of two outcomes: success or failure. The probability of success is denoted by \( p \), and the probability of failure is \( 1 - p \). If there are \( n \) trials, the binomial distribution gives the probability of having exactly \( k \) successes.

The probability mass function (PMF) of a binomial distribution is given by:
\[ P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \]
where:
- \( \binom{n}{k} = \frac{n!}{k!(n - k)!} \) is the binomial coefficient,
- \( n \) is the number of trials,
- \( k \) is the number of successes,
- \( p \) is the probability of success on a single trial.

### Given Problem

In the given problem:
- The probability of a chip receiving a thick enough coating (\( p \)) is 0.70.
- The number of chips (\( n \)) is 15.

We need to find:
(i) The probability that at least 12 chips will have thick enough coatings.
(ii) The probability that exactly 10 chips will have thick enough coatings.

### Calculations

#### (i) Probability that at least 12 chips will have thick enough coatings

To find the probability of at least 12 successes, we sum the probabilities of getting 12, 13, 14, and 15 successes.

\[ P(X \geq 12) = P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) \]

Using the binomial PMF formula:

\[ P(X = k) = \binom{15}{k} (0.70)^k (0.30)^{15 - k} \]

Let's calculate these probabilities individually:

1. \( P(X = 12) \):
\[ P(X = 12) = \binom{15}{12} (0.70)^{12} (0.30)^{3} \]
\[ P(X = 12) = \frac{15!}{12! \cdot 3!} (0.70)^{12} (0.30)^{3} \]

2. \( P(X = 13) \):
\[ P(X = 13) = \binom{15}{13} (0.70)^{13} (0.30)^{2} \]
\[ P(X = 13) = \frac{15!}{13! \cdot 2!} (0.70)^{13} (0.30)^{2} \]

3. \( P(X = 14) \):
\[ P(X = 14) = \binom{15}{14} (0.70)^{14} (0.30)^{1} \]
\[ P(X = 14) = \frac{15!}{14! \cdot 1!} (0.70)^{14} (0.30)^{1} \]

4. \( P(X = 15) \):
\[ P(X = 15) = \binom{15}{15} (0.70)^{15} (0.30)^{0} \]
\[ P(X = 15) = \frac{15!}{15! \cdot 0!} (0.70)^{15} (0.30)^{0} \]

Summing these probabilities:

\[ P(X \geq 12) = P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) \]

#### (ii) Probability that exactly 10 chips will have thick enough coatings

Using the binomial PMF formula:

\[ P(X = 10) = \binom{15}{10} (0.70)^{10} (0.30)^{5} \]
\[ P(X = 10) = \frac{15!}{10! \cdot 5!} (0.70)^{10} (0.30)^{5} \]

### Numerical Calculations

Let's perform the calculations:

1. \( P(X = 12) = \binom{15}{12} (0.70)^{12} (0.30)^{3} \)
\[ P(X = 12) = 455 \times (0.70)^{12} \times (0.30)^{3} \approx 0.196 \]

2. \( P(X = 13) = \binom{15}{13} (0.70)^{13} (0.30)^{2} \)
\[ P(X = 13) = 105 \times (0.70)^{13} \times (0.30)^{2} \approx 0.058 \]

3. \( P(X = 14) = \binom{15}{14} (0.70)^{14} (0.30)^{1} \)
\[ P(X = 14) = 15 \times (0.70)^{14} \times (0.30)^{1} \approx 0.010 \]

4. \( P(X = 15) = \binom{15}{15} (0.70)^{15} (0.30)^{0} \)
\[ P(X = 15) = 1 \times (0.70)^{15} \times (0.30)^{0} \approx 0.004 \]

Summing these probabilities:

\[ P(X \geq 12) \approx 0.196 + 0.058 + 0.010 + 0.004 = 0.268 \]

For \( P(X = 10) \):

\[ P(X = 10) = \binom{15}{10} (0.70)^{10} (0.30)^{5} \]
\[ P(X = 10) = 3003 \times (0.70)^{10} \times (0.30)^{5} \approx 0.201 \]

### Summary

(i) The probability that at least 12 chips will have thick enough coatings is approximately 0.268.

(ii) The probability that exactly 10 chips will have thick enough coatings is approximately 0.201.

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Do you want more details or have any questions? Here are some related questions you might consider:

1. What are the properties of a binomial distribution?
2. How can the binomial distribution be approximated by the normal distribution?
3. What are some applications of the binomial distribution in real-world scenarios?
4. How does changing the probability \( p \) affect the binomial distribution?
5. What is the difference between binomial and Poisson distributions?

**Tip:** When calculating probabilities for binomial distributions, ensure that the number of trials and the probability of success are correctly identified and used in the formula.

Understand the binomial distribution and calculate probabilities for the coating process of integrated circuit chips. Learn the probability of at least 12 and exactly 10 chips receiving thick enough coatings.

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