Math Problem Statement

11.A 6-sided die is rolled. What is the probability of getting a number more than 4? 12.The of Events A and B represents the number of instances in which both Events A and B occur at the same time. 14. Consider the given discrete probability distribution. P(X<=4).

X

P(x)

0

0.30

0.25

2

0.20

3

0.15

4

0.05

5

0.05

Solution

Problem 11:

For a 6-sided die, the numbers possible on each roll are: 1, 2, 3, 4, 5, and 6. We need to calculate the probability of rolling a number greater than 4.

  • Favorable outcomes (more than 4): 5 and 6.
  • Total possible outcomes: 6 (since it’s a 6-sided die).

The probability P(roll>4)P(\text{roll} > 4) is given by: P(roll>4)=number of favorable outcomestotal possible outcomes=26=13P(\text{roll} > 4) = \frac{\text{number of favorable outcomes}}{\text{total possible outcomes}} = \frac{2}{6} = \frac{1}{3}

Answer: The probability of rolling a number more than 4 is 13\frac{1}{3}.


Problem 12:

The statement provided, "The of Events A and B represents the number of instances in which both Events A and B occur at the same time," likely refers to the concept of intersection.

The intersection of Events A and B, denoted as ABA \cap B, represents the set of outcomes in which both Events A and B happen simultaneously. In probability, if P(AB)P(A \cap B) is the probability of both A and B occurring together, it is calculated based on the joint distribution of A and B.

Answer: The phrase likely refers to the intersection of events, ABA \cap B.


Problem 14:

To find P(X4)P(X \leq 4) using the given discrete probability distribution:

XP(X)
00.30
10.25
20.20
30.15
40.05
50.05

Since we want P(X4)P(X \leq 4), we sum the probabilities for X=0,1,2,3,X = 0, 1, 2, 3, and 44:

P(X4)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) =0.30+0.25+0.20+0.15+0.05=0.95= 0.30 + 0.25 + 0.20 + 0.15 + 0.05 = 0.95

Answer: The probability that X4X \leq 4 is 0.950.95.


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Follow-Up Questions:

  1. How would the probability change if we were asked about numbers greater than 3 on the die?
  2. What is the probability of either Event A or Event B occurring if they are mutually exclusive?
  3. How would the discrete probability distribution change if P(X=5)P(X=5) were adjusted to 0.10?
  4. In the case of multiple dice, how does probability of outcomes change as we increase the number of dice?
  5. How can we calculate the probability of getting any number less than 3 in this distribution?

Tip: In probability distributions, always ensure the probabilities sum to 1 for a valid distribution.

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Math Problem Analysis

Mathematical Concepts

Probability
Discrete Probability Distribution
Intersection of Events

Formulas

Probability of an event = (number of favorable outcomes) / (total possible outcomes)
Intersection of Events A and B, P(A ∩ B)
Summing probabilities for cumulative distribution, P(X ≤ k) = Σ P(X = x)

Theorems

Basic Probability Theorem
Intersection Rule for Events
Cumulative Probability Calculation

Suitable Grade Level

Grades 9-10