Question

the random variable x has the following discrete probability distribution.
x 10 11 12 13
p(x) 0.3 0.1 0.3 0.2
since the values that x can assume are mutually exclusive events, the event {x ≤ 12} is the union of three mutually exclusive events, {x = 10}∪{x = 11}∪{x = 12}.
a. find p(x ≤ 12).
p(x ≤ 12) = \square

Explanation:

Step1: Identify relevant probabilities

We need \( P(x = 10) = 0.3 \), \( P(x = 11) = 0.1 \), \( P(x = 12) = 0.3 \).

Step2: Sum the probabilities

For mutually exclusive events, \( P(A \cup B \cup C)=P(A)+P(B)+P(C) \). So \( P(x \leq 12)=P(x = 10)+P(x = 11)+P(x = 12) \).
Substitute the values: \( 0.3 + 0.1 + 0.3 = 0.7 \).

Answer:

\( 0.7 \)