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Question
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example 1: students were surveyed on campus about their study habits. some said they study in the morning, others study during the day between classes, and others study at night. some students always study in a group and others always study alone. the distribution is shown below:
how students study morning between classes evening
study in a group 7 9 4
study alone 1 6 3
if a student who was surveyed is selected at random, find these probabilities:
a.) the student studies in the evening.
b.) the student studies between classes.
c.) the student studies in the morning or in the evening.
d.) the student studies in the morning and in a group.
e.) the student studies in the evening or studies alone.
f.) the student studies between classes and alone.
Step1: Calculate total surveyed students
Add all values in the table:
$7 + 9 + 4 + 1 + 6 + 3 = 30$
Step2: Solve part (a): Evening study probability
Sum evening students, divide by total:
$\text{Probability} = \frac{4 + 3}{30} = \frac{7}{30}$
Step3: Solve part (b): Between classes probability
Sum between classes students, divide by total:
$\text{Probability} = \frac{9 + 6}{30} = \frac{15}{30} = \frac{1}{2}$
Step4: Solve part (c): Morning or evening probability
Sum morning + evening students, divide by total:
$\text{Probability} = \frac{(7+1) + (4+3)}{30} = \frac{8 + 7}{30} = \frac{15}{30} = \frac{1}{2}$
Step5: Solve part (d): Morning and group probability
Take morning group value, divide by total:
$\text{Probability} = \frac{7}{30}$
Step6: Solve part (e): Evening or alone probability
Use addition rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$\text{Probability} = \frac{7}{30} + \frac{1+6+3}{30} - \frac{3}{30} = \frac{7 + 10 - 3}{30} = \frac{14}{30} = \frac{7}{15}$
Step7: Solve part (f): Between classes and alone probability
Take between classes alone value, divide by total:
$\text{Probability} = \frac{6}{30} = \frac{1}{5}$
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a.) $\frac{7}{30}$
b.) $\frac{1}{2}$
c.) $\frac{1}{2}$
d.) $\frac{7}{30}$
e.) $\frac{7}{15}$
f.) $\frac{1}{5}$