The discrete random variable
R
R  is such that
R
∈
{
1
,
2
,
3
,
4
}
R∈{1,2,3,4}.Given that
P
(
R
=
r
)
=
k
(
r
+
1
)
r
+
2
P(R=r)=
r+2
k(r+1)
​
, then find.

(a)
The value of the constant
k
k.

(b)
P
(
R
⩽
2
)
P(R⩽2).

N
o
t
e
⁣
:
Note:Give your answer in simplest fraction form only.

To solve the problem, we need to first find the value of the constant \( k \) and then compute \( P(R \leq 2) \).

### (a) Finding the constant \( k \)

Given the probability mass function:
\[ P(R = r) = \frac{k(r + 1)}{r + 2} \]
where \( R \in \{1, 2, 3, 4\} \), we know that the sum of all probabilities must equal 1:
\[ \sum_{r=1}^{4} P(R = r) = 1 \]

Let's compute the individual probabilities:
\[ P(R = 1) = \frac{k(1 + 1)}{1 + 2} = \frac{2k}{3} \]
\[ P(R = 2) = \frac{k(2 + 1)}{2 + 2} = \frac{3k}{4} \]
\[ P(R = 3) = \frac{k(3 + 1)}{3 + 2} = \frac{4k}{5} \]
\[ P(R = 4) = \frac{k(4 + 1)}{4 + 2} = \frac{5k}{6} \]

Now, sum these probabilities:
\[ \frac{2k}{3} + \frac{3k}{4} + \frac{4k}{5} + \frac{5k}{6} = 1 \]

To combine these fractions, find a common denominator. The least common multiple of 3, 4, 5, and 6 is 60. Convert each fraction:
\[ \frac{2k}{3} = \frac{40k}{60} \]
\[ \frac{3k}{4} = \frac{45k}{60} \]
\[ \frac{4k}{5} = \frac{48k}{60} \]
\[ \frac{5k}{6} = \frac{50k}{60} \]

Summing these:
\[ \frac{40k}{60} + \frac{45k}{60} + \frac{48k}{60} + \frac{50k}{60} = 1 \]
\[ \frac{183k}{60} = 1 \]

Solving for \( k \):
\[ k = \frac{60}{183} = \frac{20}{61} \]

So, the value of the constant \( k \) is:
\[ k = \frac{20}{61} \]

### (b) Finding \( P(R \leq 2) \)

\[ P(R \leq 2) = P(R = 1) + P(R = 2) \]

Using the probabilities calculated with \( k = \frac{20}{61} \):
\[ P(R = 1) = \frac{2k}{3} = \frac{2 \times \frac{20}{61}}{3} = \frac{40}{183} \]
\[ P(R = 2) = \frac{3k}{4} = \frac{3 \times \frac{20}{61}}{4} = \frac{60}{244} = \frac{30}{122} = \frac{15}{61} \]

Sum these probabilities:
\[ P(R \leq 2) = \frac{40}{183} + \frac{15}{61} \]

Convert \(\frac{15}{61}\) to have the same denominator:
\[ \frac{15}{61} = \frac{45}{183} \]

Thus:
\[ P(R \leq 2) = \frac{40}{183} + \frac{45}{183} = \frac{85}{183} \]

In simplest form, this is already simplified. Therefore:
\[ P(R \leq 2) = \frac{85}{183} \]

To summarize:
(a) The value of the constant \( k \) is \( \frac{20}{61} \).
(b) \( P(R \leq 2) = \frac{85}{183} \).

Do you want details or have any questions?

Here are 5 numbered questions for you to ask:

1. How do you calculate probabilities for discrete random variables?
2. Can you explain the steps to find the constant in a probability distribution?
3. How do you find the common denominator for adding fractions?
4. What are some tips for simplifying fractions?
5. How can you check if your probability calculations are correct?

**Tip:** When dealing with fractions, always look for the least common multiple to simplify the addition or subtraction of fractions effectively.

Learn how to find the constant k and compute the probability P(R <= 2) for a discrete random variable R with a given probability mass function. This problem covers key concepts in probability theory and discrete random variables, suitable for undergraduate students.

Math Problem Statement

The discrete random variable R R is such that R ∈ { 1 , 2 , 3 , 4 } R∈{1,2,3,4}.Given that P ( R

r )

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