Question

Number of Absences
The monthly absences of a learner based on his class adviser's record are
presented in the probability distribution below:
X
(Number of Boxes ) 0
1
2
3
4
1
3
P(X)
3
1
10
(Probability )
10 10 5
1
10
1. What is the probability that the number of absences is more than 3?
2. What is the probability that the number of absences is at least 2?
3. What is the probability that the number of absences is greater than 1 but
less than 4?​

Answer 1

Answer:

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Answer 2

Answer:

(a) Probability that number of absences are more than 3 = $\frac{1}{5}$

(b)Probability that number of absences are at least 2 = $\frac{6}{10}$

(c)Probability that number of absences is greater than 1 but less than 4 =$\frac{2}{5}$

Step-by-step explanation:

Step 1 of 3

(a)Probability that number of absences are more than 3

$P(X > 3)=P(X=4)\\P(X > 3)=\frac{1}{5}$

Step 2 of 3

(b)Probability that number of absences are at least 2

The word at least states that 2 and more than 2 but less than two are not accepted.

$P(X\geq 2)=P(X=2)+P(X=3)+P(X=4)\\P(X\geq 2)=\frac{3}{10}+ \frac{1}{10} +\frac{1}{5}\\P(X\geq 2)=\frac{3}{5}$

Step 3 of 3

(c) Probability that the number of absences is greater than 1 but less than 4

These are all the probabilities that lie between 1 and 4, so we get:

$P(1 < X < 4)=P(X=2)+P(X=3)\\P(1 < X < 4)=\frac{3}{10} + \frac{1}{10}\\P(1 < X < 4)=\frac{4}{10}=\frac{2}{5}$

So we get the answers as:

(a) Probability that number of absences are more than 3 = $\frac{1}{5}$

(b)Probability that number of absences are at least 2 = $\frac{6}{10}$

(c)Probability that number of absences is greater than 1 but less than 4 =$\frac{2}{5}$