To solve each of your questions, we need to set up the probabilities correctly based on the problem statement.
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What is the probability that Amy gives out a picture on the first day?
The probability that Amy goes out on the first day is \( p \), and the probability that someone is at the house (after she decides to go out) is \( q \). Therefore, the probability that she gives out a picture on the first day (goes out and someone is there) is: \[ P(\text{gives out picture on day 1}) = p \cdot q \]
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What is the probability that she gave out exactly k pictures on the first n days?
We can treat each day's event of giving out a picture as a Bernoulli trial: she has a probability \( p \cdot q \) of giving out a picture and a probability \( 1 - pq \) of not giving it out. The random variable for the number of pictures given out follows a binomial distribution:
Thus, the probability that she gives out exactly k pictures in n days is given by: \[ P(X = k) = \binom{n}{k} (pq)^k (1 - pq)^{n-k} \]
Where \( \binom{n}{k} \) is the binomial coefficient, representing "n choose k".
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What is the probability that she stayed home on day 1, given that she did not give out a picture on that day?
We need to use Bayes' theorem for this conditional probability. Define:
- \( H \): the event that Amy stayed home on day 1.
- \( N \): the event that Amy did not give out a picture on day 1.
We want to find \( P(H | N) \).
By Bayes' theorem, we have: \[ P(H | N) = \frac{P(N | H)P(H)}{P(N)} \]
Where:
- \( P(H) = 1 - p \) (the probability she stayed home).
- \( P(N | H) = 1 \) (if she stayed home she definitely did not give out a picture).
- \( P(N) = P(N | H) P(H) + P(N | G) P(G) \)
- \( G \): the event that she goes out on day 1.
- \( P(N | G) = (1-q) \).
- \( P(G) = p \).
Therefore, \[ P(N) = (1)(1 - p) + (1 - q)p = (1 - p) + p(1 - q) = 1 - p + p - pq = 1 - pq \]
Putting it into Bayes' theorem, \[ P(H | N) = \frac{(1)(1 - p)}{1 - pq} = \frac{1 - p}{1 - pq} \]
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Within the first week (7 days), what is the probability that there were no days in which Amy went out but did not give out a picture of her hometown?
The event "she went out but did not give out a picture" happens if she goes out (with probability \( p \)) and no one is home (with probability \( 1 - q \)). The probability of this happening on any given day is \( p(1 - q) \).
We are interested in the probability that this does not happen at all in a week (7 days). Therefore, the probability that she either stays home or gives out a picture on any given day is: \[ 1 - p(1 - q) \]
For 7 days, the probability that this condition is not met any day is: \[ \left(1 - p(1 - q)\right)^7 \]
Therefore, the final answer for this part is: \[ P(\text{no days went out but did not give out}) = (1 - p(1 - q))^7 \]
Summarizing the answers:
- \( p \cdot q \)
- \( \binom{n}{k} (pq)^k (1 - pq)^{n-k} \)
- \( \frac{1 - p}{1 - pq} \)
- \( (1 - p(1 - q))^7 \)