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.
- 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: \[ P(X \geq 3) = P(3) + P(4) + P(5) + P(6) + P(7) + P(8) + P(9) + P(10) = 8 \times 0.1 = 0.8. \]
- 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:
- \( P(X \geq 3) = 0.8 \)
- \( P(X \leq 4) = 0.4 \)
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