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The problem in the image involves the binomial distribution, where the probability that a married woman dislikes her mother-in-law is 85% (or \( p = 0.85 \)) and we are analyzing a group of 12 married women (\( n = 12 \)).

### The binomial formula:
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]
Where:
- \( n = 12 \) is the number of trials (women),
- \( p = 0.85 \) is the probability of success (dislike),
- \( k \) is the number of women who dislike their mother-in-law.

Let's go through the parts:

---

### (a) All of them dislike their mother-in-law.

We need to find \( P(X = 12) \), where \( X \) is the number of women who dislike their mother-in-law.

\[
P(X = 12) = \binom{12}{12} (0.85)^{12} (0.15)^{0}
\]
\[
P(X = 12) = 1 \cdot (0.85)^{12} \approx 0.148
\]

So, the probability that all of them dislike their mother-in-law is approximately **0.148**.

---

### (b) None of them dislike their mother-in-law.

This is already given as \( P(X = 0) = 0.000003 \).

---

### (c) At least ten of them dislike their mother-in-law.

We need to calculate \( P(X \geq 10) \), which is the sum of \( P(X = 10) \), \( P(X = 11) \), and \( P(X = 12) \).

1. \( P(X = 10) \):
\[
P(X = 10) = \binom{12}{10} (0.85)^{10} (0.15)^2
\]
\[
P(X = 10) \approx 66 \cdot (0.85)^{10} \cdot (0.15)^2 \approx 66 \cdot 0.198 \cdot 0.0225 \approx 0.293
\]

2. \( P(X = 11) \):
\[
P(X = 11) = \binom{12}{11} (0.85)^{11} (0.15)^1
\]
\[
P(X = 11) = 12 \cdot (0.85)^{11} \cdot (0.15) \approx 12 \cdot 0.233 \cdot 0.15 \approx 0.420
\]

3. From part (a), \( P(X = 12) = 0.148 \).

Thus,
\[
P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12) = 0.293 + 0.420 + 0.148 = 0.861
\]

So, the probability that at least 10 of them dislike their mother-in-law is approximately **0.861**.

---

### (d) No more than nine of them dislike their mother-in-law.

We need to calculate \( P(X \leq 9) \), which is \( 1 - P(X \geq 10) \) (since the total probability is 1).

\[
P(X \leq 9) = 1 - P(X \geq 10) = 1 - 0.861 = 0.139
\]

So, the probability that no more than 9 of them dislike their mother-in-law is approximately **0.139**.

---

Would you like more details on any part of the solution?

### Here are 5 related questions to expand your understanding:
1. How does the binomial distribution change if only 10 women were surveyed instead of 12?
2. What is the probability that exactly 8 women dislike their mother-in-law in this scenario?
3. How would you calculate the mean and standard deviation for this binomial distribution?
4. What if the dislike probability dropped to 70%—how would that affect the answers?
5. Can you calculate the cumulative probability for \( P(X \leq 3) \) in this case?

### Tip:
When calculating probabilities for "at least" or "no more than" situations, summing the individual probabilities or using complements can save time.

Sociologists say that 85% of married women claim that their husband's mother is the biggest bone of contention in their marriages. Suppose that twelve married women are having coffee together one morning. Find the following probabilities:
(a) All of them dislike their mother-in-law.
(b) None of them dislike their mother-in-law.
(c) At least ten of them dislike their mother-in-law.
(d) No more than nine of them dislike their mother-in-law.

Explore how to calculate the probabilities using binomial distribution where 85% of women dislike their mother-in-law. This problem involves finding probabilities of all, none, at least ten, and no more than nine women disliking their mother-in-law out of a group of twelve.

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