Math Problem Statement

y P(Y=y) 0 0.187 1 0.274 2 0.219 3 0.094 4 0.042 5 0.043 6 0.021 7 0.013 8 0.107 d. Use the special addition rule and the probability distribution to determine the probability that the year had at least one major hurricane. The probability is    enter your response here. ​(Type an integer or decimal rounded to three decimal places as​ needed.) Part 5 e. Use the special addition rule and the probability distribution to determine the probability that the year had exactly two major hurricanes. The probability is    enter your response here. ​(Type an integer or decimal rounded to three decimal places as​ needed.) Part 6 f. Use the special addition rule and the probability distribution to determine the probability that the year had between 3 and 5 major​ hurricanes, inclusive. The probability is    enter your response here. ​(Type an integer or decimal rounded to three decimal places as​ needed.) Complete parts (a) through (f) below.Complete parts (a) through (f) below.Select the correct choice below and fill in the answer box within your choice.input field 3input field 4Select the correct choice below and fill in the answer box within your choice.input field 5input field 6Select the correct choice below and fill in the answer box within your choice.The probability is(Type an integer or decimal rounded to three decimal places as needed.)The probability is(Type an integer or decimal rounded to three decimal places as needed.)The probability is(Type an integer or decimal rounded to three decimal places as needed.)

Solution

Let's go through each part systematically.

Part d: Probability that the year had at least one major hurricane

To find the probability that there was at least one major hurricane, we need to use the complement rule. The complement rule states that the probability of at least one event occurring is P(at least one hurricane)=1P(no hurricanes)P(\text{at least one hurricane}) = 1 - P(\text{no hurricanes}).

From the table, the probability that the year had no hurricanes is given as P(Y=0)=0.187P(Y = 0) = 0.187.

Thus, the probability of at least one major hurricane is: P(at least one hurricane)=10.187=0.813P(\text{at least one hurricane}) = 1 - 0.187 = 0.813

So, the probability is: 0.813\boxed{0.813}

Part e: Probability that the year had exactly two major hurricanes

From the table, the probability that the year had exactly two major hurricanes is: P(Y=2)=0.219P(Y = 2) = 0.219

So, the probability is: 0.219\boxed{0.219}

Part f: Probability that the year had between 3 and 5 major hurricanes (inclusive)

To find the probability that the year had between 3 and 5 major hurricanes, inclusive, we need to sum the probabilities for Y=3Y = 3, Y=4Y = 4, and Y=5Y = 5.

From the table:

  • P(Y=3)=0.094P(Y = 3) = 0.094
  • P(Y=4)=0.042P(Y = 4) = 0.042
  • P(Y=5)=0.043P(Y = 5) = 0.043

Thus, the total probability is: P(3Y5)=0.094+0.042+0.043=0.179P(3 \leq Y \leq 5) = 0.094 + 0.042 + 0.043 = 0.179

So, the probability is: 0.179\boxed{0.179}


Let me know if you would like further details or explanations!

Here are 5 related questions:

  1. What is the total probability for the given distribution?
  2. What is the probability of having more than 5 major hurricanes in a year?
  3. What is the probability of having fewer than 2 major hurricanes in a year?
  4. How would the complement rule apply if we wanted to calculate "no more than 2 hurricanes"?
  5. What is the expected number of major hurricanes in a given year?

Tip: For probabilities involving ranges, simply add the individual probabilities within the specified range!

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Math Problem Analysis

Mathematical Concepts

Probability
Addition Rule
Complement Rule

Formulas

P(at least one event) = 1 - P(no event)
P(exactly k events) = P(Y=k)
P(a ≤ Y ≤ b) = P(Y=a) + P(Y=b)

Theorems

Addition Rule of Probability
Complement Rule of Probability

Suitable Grade Level

Grades 10-12