To solve the problems, we will break them down step by step according to the described scenarios.
1. Probability that Amy gives out a picture on the first day
On the first day, two things need to happen for Amy to give out a picture:
- She must decide to go out, which happens with probability \( p \).
- Someone must be home when she visits the house, which happens with probability \( q \).
The probability that Amy gives out a picture on the first day is thus:
\[ P(\text{gives out a picture on day 1}) = p \times q \]
2. Probability that she gave out exactly \( k \) pictures in the first \( n \) days
To compute the probability that Amy gives out exactly \( k \) pictures over \( n \) days, we can model this situation as a binomial experiment, where:
- Each day she has two possible outcomes: giving out a picture (success) or not (failure).
- The probability of a "success" (giving out a picture) on any given day is \( p \times q \) and the probability of a "failure" is \( 1 - (p \times q) \).
This can be described by the binomial probability formula:
\[ P(X = k) = \binom{n}{k} (p \times q)^k (1 - (p \times q))^{n-k} \]
Here, \( \binom{n}{k} \) is the binomial coefficient, representing the number of ways to choose \( k \) days out of \( n \) to be successful.
3. Probability that she stayed home on day 1 given that she did not give out a picture on that day
We want to find \( P(\text{stayed home on day 1} | \text{did not give out a picture on day 1}) \).
Using Bayes' theorem:
\[ P(A|B) = \frac{P(B|A)P(A)}{P(B)} \]
Let \( A \) be the event that Amy stayed home on day 1, and \( B \) be the event that she did not give out a picture on day 1.
From the problem statement, we have:
- \( P(A) = 1 - p \) (the probability she stayed home),
- \( P(B|A) = 1 \) (if she stayed home, she definitely did not give out a picture),
- \( P(B) \) is the total probability that she does not give out a picture on day 1, which consists of two parts:
- She stayed home: \( (1 - p) \)
- She went out but no one was home: \( p \times (1 - q) \)
Thus:
\[ P(B) = (1 - p) + p \times (1 - q) = 1 - p + p - pq = 1 - pq \]
Now, we can compute:
\[ P(A|B) = \frac{1 \cdot (1 - p)}{1 - pq} \]
4. Probability that there were no days in which Amy went out but did not give out a picture of her hometown within the first week
For Amy to go out but not give out a picture on a given day, it means:
- She goes out, which happens with probability \( p \).
- No one is home at the house, which happens with probability \( (1 - q) \).
The probability of this happening on a single day is:
\[ P(\text{went out but did not give picture}) = p \times (1 - q) \]
Now, we want the probability that this event does not occur on any of the 7 days. The probability that it does not happen on a single day is:
\[ 1 - P(\text{went out but did not give picture}) = 1 - (p \times (1 - q)) \]
Since we are considering 7 days, under the assumption that the events are independent, the probability that this event does not occur on any of the 7 days is:
\[ P(\text{no days went out but did not give picture}) = [1 - p(1 - q)]^7 \]
In summary:
- \( p \cdot q \)
- \( \binom{n}{k} (p \cdot q)^k (1 - (p \cdot q))^{n-k} \)
- \( \frac{(1 - p)}{(1 - pq)} \)
- \( [1 - p(1 - q)]^7 \)
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