Let's break down each question and solve it step by step.

Given probabilities:

• $P(\text{miss 1 day}) = 0.24$
• $P(\text{miss 2 days}) = 0.14$
• $P(\text{miss 3 or more days}) = 0.27$

The total probability of missing any school day: $P(\text{miss 1 or more days}) = 0.24 + 0.14 + 0.27 = 0.65$ Thus, the probability of missing no days of school: $P(\text{miss 0 days}) = 1 - P(\text{miss 1 or more days}) = 1 - 0.65 = 0.35$

1. What is the probability that a student chosen at random doesn't miss any days of school due to sickness this year?

$P(\text{miss 0 days}) = 0.35$ Answer: 0.35

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2. What is the probability that a student chosen at random misses no more than one day?

This is the sum of the probabilities of missing 0 or 1 day: $P(\text{miss 0 or 1 day}) = P(\text{miss 0 days}) + P(\text{miss 1 day}) = 0.35 + 0.24 = 0.59$ Answer: 0.59

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3. What is the probability that a student chosen at random misses at least one day?

$P(\text{miss 1 or more days}) = 1 - P(\text{miss 0 days}) = 1 - 0.35 = 0.65$ Answer: 0.65

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4. If a parent has two kids at an elementary school in this county, what is the probability that neither kid will miss any school? Assume the probabilities are independent.

For each child, the probability of missing no days is $P(\text{miss 0 days}) = 0.35$. Since the events are independent, the probability that both children will miss no days is: $P(\text{both miss 0 days}) = P(\text{miss 0 days}) \times P(\text{miss 0 days}) = 0.35 \times 0.35 = 0.1225$ Answer: 0.1225

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5. If a parent has two kids at an elementary school in this county, what is the probability that both kids will miss some school, i.e., at least one day? Assume the probabilities are independent.

For each child, the probability of missing at least one day is $P(\text{miss 1 or more days}) = 0.65$. Since the events are independent, the probability that both children miss at least one day is: $P(\text{both miss some days}) = P(\text{miss 1 or more days}) \times P(\text{miss 1 or more days}) = 0.65 \times 0.65 = 0.4225$ Answer: 0.4225

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Summary of answers:

1. $P(\text{miss 0 days}) = 0.35$
2. $P(\text{miss 0 or 1 days}) = 0.59$
3. $P(\text{miss 1 or more days}) = 0.65$
4. $P(\text{both miss 0 days}) = 0.1225$
5. $P(\text{both miss some days}) = 0.4225$

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Would you like further details on any of the solutions or have any specific questions?

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Here are 5 relative questions:

1. What is the probability that at least one of the two kids will miss no school?
2. How would the answers change if the probability of missing 0 days was 0.4 instead of 0.35?
3. What is the probability that one kid misses exactly 2 days while the other misses no days?
4. How do you calculate the probability that exactly one kid misses some school while the other misses no school?
5. If three kids are involved, how would the probability of all three missing at least one day be calculated?

Tip: When calculating probabilities of independent events, always multiply the individual probabilities.