Math Problem Statement

Use the extended multiplication rule to calculate the following probabilities. Round your answers to 3 digits.

(a) If you flip a fair coin 5 times, what is the probability of getting 5 heads? Blank 1

(b) If you randomly select 3 people, what is the probability that they were born on the same day of the week (Monday, Tuesday, ... Sunday)? Blank 2

(c) Assume that your car starts 99% of the time. What is the probability that your car will start for the next 14 days in a row? Assume these events are independent. Blank 3

Solution

Let's break down each of these probability problems using the extended multiplication rule, which states that the probability of a series of independent events happening is the product of the individual probabilities.


(a) Probability of getting 5 heads when flipping a fair coin 5 times.

  • A fair coin has two outcomes: heads (H) and tails (T), each with a probability of 12\frac{1}{2}.
  • The flips are independent, so we multiply the probabilities for each flip.

The probability of getting heads on each flip is 12\frac{1}{2}. For 5 heads in a row, the probability is: P(5 heads)=(12)5=132=0.03125P(\text{5 heads}) = \left(\frac{1}{2}\right)^5 = \frac{1}{32} = 0.03125 Rounded to 3 digits:
Blank 1 = 0.031


(b) Probability that 3 randomly selected people were born on the same day of the week.

  • There are 7 days in a week, so the probability that the first person was born on any particular day is 1 (it's certain).
  • The second person must be born on the same day, so the probability for them is 17\frac{1}{7}.
  • The third person must also be born on that same day, so their probability is also 17\frac{1}{7}.

The total probability is: P(same day)=1×17×17=1490.0204P(\text{same day}) = 1 \times \frac{1}{7} \times \frac{1}{7} = \frac{1}{49} \approx 0.0204 Rounded to 3 digits:
Blank 2 = 0.020


(c) Probability that the car will start for 14 days in a row, given that the car starts 99% of the time.

  • The probability of the car starting on any given day is 0.990.99.
  • Since the events are independent, we multiply the probability for each of the 14 days.

The total probability is: P(car starts 14 days)=0.99140.868P(\text{car starts 14 days}) = 0.99^{14} \approx 0.868 Rounded to 3 digits:
Blank 3 = 0.868


Summary of answers:

  • Blank 1 = 0.031
  • Blank 2 = 0.020
  • Blank 3 = 0.868

Would you like more detailed explanations or have any questions about the calculations?

Further Related Questions:

  1. What would the probability be if you flipped the coin 10 times and wanted 10 heads?
  2. What is the probability that at least 2 people out of 3 were born on the same day of the week?
  3. What is the probability that your car will fail to start at least once in 14 days?
  4. How would the probabilities change if you were selecting 5 people instead of 3?
  5. How do you calculate the probability for events that are not independent?

Tip: When dealing with repeated independent events, remember that the extended multiplication rule is your best tool to calculate the probability for all events happening together.

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Math Problem Analysis

Mathematical Concepts

Probability
Independent Events
Multiplication Rule

Formulas

P(A and B) = P(A) * P(B)
P(A) = p^n (for n independent events)

Theorems

Extended Multiplication Rule
Independent Events Probability

Suitable Grade Level

Grades 9-11