A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN: 9780134753119
Author: Sheldon Ross
Publisher: PEARSON
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I need help with this question 9

Use the probability distribution for the random variable \( x \) to answer the question.

\[
\begin{array}{c|ccccc}
x & 0 & 1 & 2 & 3 & 4 \\
\hline
p(x) & 0.12 & 0.2 & 0.2 & 0.36 & 0.12 \\
\end{array}
\]

What is the probability that \( x \) is 3 or less?

**Solution:**

To find the probability that \( x \) is 3 or less, sum the probabilities of \( x \) taking the values 0, 1, 2, and 3.

\[
P(x \leq 3) = p(0) + p(1) + p(2) + p(3) = 0.12 + 0.2 + 0.2 + 0.36
\]

Calculate the sum:

\[
P(x \leq 3) = 0.88
\]

Therefore, the probability that \( x \) is 3 or less is 0.88.
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Transcribed Image Text:Use the probability distribution for the random variable \( x \) to answer the question. \[ \begin{array}{c|ccccc} x & 0 & 1 & 2 & 3 & 4 \\ \hline p(x) & 0.12 & 0.2 & 0.2 & 0.36 & 0.12 \\ \end{array} \] What is the probability that \( x \) is 3 or less? **Solution:** To find the probability that \( x \) is 3 or less, sum the probabilities of \( x \) taking the values 0, 1, 2, and 3. \[ P(x \leq 3) = p(0) + p(1) + p(2) + p(3) = 0.12 + 0.2 + 0.2 + 0.36 \] Calculate the sum: \[ P(x \leq 3) = 0.88 \] Therefore, the probability that \( x \) is 3 or less is 0.88.
Expert Solution
Check Mark
Step 1

Pmf given

X 0 1 2 3 4
P(X=x) 0.12 0.2

0.20

0.36 0.12
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