Question

A random variable X assumes values 1,2,3,..., 8,9,and 10, each with the same probability, namely the probability 0.1.

1) Find the probability of X having value of at least 3.

2) Find the probability of X getting no more than 4.

Answer 1

To solve the problems regarding the random variable \( X \) that assumes values from the set \( \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \) with equal probabilities of \( 0.1 \) for each value, we can use the definition of probabilities.

1) **Find the probability of \( X \) having a value of at least 3.**

The values of \( X \) that are at least 3 are \( \{3, 4, 5, 6, 7, 8, 9, 10\} \).

To find the probability:
- Count the favorable outcomes: There are 8 outcomes (3 through 10).
- Each outcome has a probability of 0.1.

So, the probability of \( X \) being at least 3 is:
\[
P(X \geq 3) = P(3) + P(4) + P(5) + P(6) + P(7) + P(8) + P(9) + P(10) = 8 \times 0.1 = 0.8.
\]

2) **Find the probability of \( X \) getting no more than 4.**

The values of \( X \) that are no more than 4 are \( \{1, 2, 3, 4\} \).

To find the probability:
- Count the favorable outcomes: There are 4 outcomes (1 through 4).
- Each outcome has a probability of 0.1.

So, the probability of \( X \) being no more than 4 is:
\[
P(X \leq 4) = P(1) + P(2) + P(3) + P(4) = 4 \times 0.1 = 0.4.
\]

In summary:
1. \( P(X \geq 3) = 0.8 \)
2. \( P(X \leq 4) = 0.4 \)