Math Problem Statement

Suppose X is uniformly distributed over [0, 5] and Y is uniformly distributed over [0, 4]. If X and Y are independent, then P(max(X, Y) > 3) is...

Solution

It looks like you've uploaded a math problem involving uniform distributions and probability. Based on the image you shared, the question is:

Suppose XX is uniformly distributed over [0,5][0, 5] and YY is uniformly distributed over [0,4][0, 4]. If XX and YY are independent, then P(max(X,Y)>3)P(\max(X, Y) > 3) is...

Let's solve this step by step:

  1. The probability of XX and YY being independent means that we can multiply their individual probabilities.
  2. Since XUniform(0,5)X \sim \text{Uniform}(0, 5) and YUniform(0,4)Y \sim \text{Uniform}(0, 4), we know their probability density functions (PDFs):
    • The PDF for XX, fX(x)=15f_X(x) = \frac{1}{5} for x[0,5]x \in [0, 5].
    • The PDF for YY, fY(y)=14f_Y(y) = \frac{1}{4} for y[0,4]y \in [0, 4].

Now, the problem asks for the probability that the maximum of XX and YY is greater than 3, i.e., P(max(X,Y)>3)P(\max(X, Y) > 3).

This is the complement of the event where both XX and YY are less than or equal to 3. Hence: P(max(X,Y)>3)=1P(X3,Y3)P(\max(X, Y) > 3) = 1 - P(X \leq 3, Y \leq 3)

Now, we calculate P(X3,Y3)P(X \leq 3, Y \leq 3) under the assumption that XX and YY are independent: P(X3,Y3)=P(X3)P(Y3)P(X \leq 3, Y \leq 3) = P(X \leq 3) \cdot P(Y \leq 3)

For XUniform(0,5)X \sim \text{Uniform}(0, 5), P(X3)=35P(X \leq 3) = \frac{3}{5}.

For YUniform(0,4)Y \sim \text{Uniform}(0, 4), P(Y3)=34P(Y \leq 3) = \frac{3}{4}.

Thus: P(X3,Y3)=35×34=920P(X \leq 3, Y \leq 3) = \frac{3}{5} \times \frac{3}{4} = \frac{9}{20}

Finally, the probability we need is: P(max(X,Y)>3)=1920=1120P(\max(X, Y) > 3) = 1 - \frac{9}{20} = \frac{11}{20}

So the probability is 1120\frac{11}{20}.

Let me know if you'd like further clarification!


Here are 5 related questions for further exploration:

  1. How can you calculate probabilities when random variables are dependent?
  2. What happens to this probability if the distributions of XX and YY change?
  3. How do you calculate the cumulative distribution function (CDF) for uniform distributions?
  4. How would you approach a similar problem if XX and YY were normally distributed?
  5. What is the relationship between the mean of a uniform distribution and the range of values?

Tip: When calculating probabilities involving the maximum or minimum of two independent random variables, it's often helpful to use the complement rule, as shown in this solution!

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Math Problem Analysis

Mathematical Concepts

Uniform Distribution
Probability
Independence of Random Variables
Complement Rule

Formulas

P(max(X, Y) > 3) = 1 - P(X ≤ 3, Y ≤ 3)
P(X ≤ a) = a/b for Uniform(a, b)

Theorems

Complement Rule for Probability
Independence of Events

Suitable Grade Level

Grades 10-12