Question
Following is the probability distribution of a random variable that represents the number of extracurricular activities a college freshman participates in.
x| 0 1 2 3 4
P(x)| 0.05 0.14 0.45 0.23 0.13
(a) Find the probability that a student participates in exactly three activities.
x| 0 1 2 3 4
P(x)| 0.05 0.14 0.45 0.23 0.13
(a) Find the probability that a student participates in exactly three activities.
Answers
Answer
(b) Find the probability that a student participates in less than three activities.
Answer
(c) Find the probability that a student participates in at least two activities.
Answered by
GPT-4o mini
To find the probability that a student participates in exactly three activities, we can refer directly to the given probability distribution.
From the provided information, we can see:
\[
P(3) = 0.23
\]
Thus, the probability that a student participates in exactly three activities is **0.23**.
From the provided information, we can see:
\[
P(3) = 0.23
\]
Thus, the probability that a student participates in exactly three activities is **0.23**.
Answered by
GPT-4o mini
To find the probability that a student participates in less than three activities, we need to sum the probabilities of the student participating in 0, 1, or 2 activities.
From the probability distribution:
- \( P(0) = 0.05 \)
- \( P(1) = 0.14 \)
- \( P(2) = 0.45 \)
Now, we sum these probabilities:
\[
P(X < 3) = P(0) + P(1) + P(2)
\]
Calculating this gives:
\[
P(X < 3) = 0.05 + 0.14 + 0.45 = 0.64
\]
Therefore, the probability that a student participates in less than three activities is **0.64**.
From the probability distribution:
- \( P(0) = 0.05 \)
- \( P(1) = 0.14 \)
- \( P(2) = 0.45 \)
Now, we sum these probabilities:
\[
P(X < 3) = P(0) + P(1) + P(2)
\]
Calculating this gives:
\[
P(X < 3) = 0.05 + 0.14 + 0.45 = 0.64
\]
Therefore, the probability that a student participates in less than three activities is **0.64**.
Answered by
GPT-4o mini
To find the probability that a student participates in at least two activities, we need to consider the probabilities of participating in 2, 3, or 4 activities.
From the probability distribution, we have:
- \( P(2) = 0.45 \)
- \( P(3) = 0.23 \)
- \( P(4) = 0.13 \)
Now, we sum these probabilities:
\[
P(X \geq 2) = P(2) + P(3) + P(4)
\]
Calculating this gives:
\[
P(X \geq 2) = 0.45 + 0.23 + 0.13 = 0.81
\]
Therefore, the probability that a student participates in at least two activities is **0.81**.
From the probability distribution, we have:
- \( P(2) = 0.45 \)
- \( P(3) = 0.23 \)
- \( P(4) = 0.13 \)
Now, we sum these probabilities:
\[
P(X \geq 2) = P(2) + P(3) + P(4)
\]
Calculating this gives:
\[
P(X \geq 2) = 0.45 + 0.23 + 0.13 = 0.81
\]
Therefore, the probability that a student participates in at least two activities is **0.81**.
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