To solve the questions step-by-step, let's outline the probabilities and assumptions provided.
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Probability that Amy gives out a picture on the first day
To give out a picture, Amy must go out (which happens with probability \( p \)) and someone must be at the house (which happens with probability \( q \)). Thus, the probability \( P(\text{gives out a picture on day 1}) \) can be given as: \[ P(\text{gives out a picture on day 1}) = p \times q \]
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Probability that she gave out exactly \( k \) pictures on the first \( n \) days
For Amy to give out exactly \( k \) pictures in \( n \) days, she must go out \( n \) times (in exactly \( k \) of those visits someone must be home) and stay home on \( n - k \) days. This follows a binomial distribution:
- The number of days Amy goes out follows a binomial distribution \( B(n, p) \).
- The number of times out of those \( n \) that someone is home (and she gives out a picture) follows another binomial distribution \( B(k, q) \).
The total probability, therefore, can be expressed using the binomial probability formula: \[ P(\text{exactly } k \text{ pictures in } n \text{ days}) = \binom{n}{k} p^n q^k (1 - q)^{n-k} \] where \( \binom{n}{k} \) is the binomial coefficient which can be expressed as \( (fact(n) / (fact(k) \cdot fact(n - k))) \).
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Probability that she stayed home on day 1, given she did not give out a picture on that day
We need to apply Bayes' theorem here. We need to find \( P(\text{stayed home on day 1} | \text{did not give out a picture}) \).
- Let \( S \) denote the event that Amy stayed home on day 1.
- Let \( N \) denote the event that Amy did not give out a picture on day 1.
We want \( P(S | N) \), which is given by: \[ P(S | N) = \frac{P(N | S) \cdot P(S)}{P(N)} \]
- \( P(S) = 1 - p \)
- If she stayed home, she can't give out a picture, hence \( P(N | S) = 1 \).
- Now we need \( P(N) \), which includes two scenarios:
- Staying home: \( P(N | S) = 1 \) with probability \( 1 - p \)
- Going out but no one home: \( P(N | \text{went out}) = 1 - q \) with probability \( p \)
So: \[ P(N) = P(N | S) \cdot P(S) + P(N | \text{went out}) \cdot P(\text{went out}) = 1 \cdot (1 - p) + (1 - q) \cdot p \] Thus, \[ P(N) = (1 - p) + p(1 - q) \] Putting it all together: \[ P(S | N) = \frac{(1) \cdot (1 - p)}{(1 - p) + p(1 - q)} \]
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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 (7 days)
We want to calculate the probability of the complementary event \( K \), where on any day she goes out without giving away a picture. On a day she goes out, if no one is home, she does not give a picture.
The probability that Amy went out and no one was home is \( p(1 - q) \).
We want the probability that none of the 7 days fall into this category where she goes out but does not give a picture, which follows:
The probability of complying with \( K' \) (not going out without giving a picture) on each day is:
- Stays home: \( 1 - p \)
- Goes out and gives a picture: \( pq \)
Thus, \[ P(\text{none of the days go out and not give picture}) = (1 - p(1 - q))^7 \] Simplifying this gives the required probability.
Let me know if you need any further clarifications or assistance!