Question

the two - way table summarizes whether or not a soccer team had practice when it was raining and when it was not raining at the end of the school day. are the events of \holding soccer practice\ and
aining at the end of the school day\ dependent or independent events? select the best answer and explanation below.

	soccer practice	no soccer practice
not raining	14	2

a) they appear to be independent events because
$p(\text{holding soccer practice }|\text{ raining}) = 0.70$ and $p(\text{holding soccer practice}) = 0.81$ and the values are not close to each other.
b) they appear to be dependent events because
$p(\text{holding soccer practice }|\text{ raining}) = 0.70$ and $p(\text{holding soccer practice}) = 0.81$ and the values are not close to each other.
c) they appear to be independent events because
$p(\text{holding soccer practice}) \
eq p(\text{raining})$.
d) they appear to be dependent events because
$p(\text{raining }|\text{ holding soccer practice}) = 0.33$ and $p(\text{raining}) = 0.38$ are relatively close together.

Explanation:

To determine if two events \( A \) (holding soccer practice) and \( B \) (raining) are independent, we check if \( P(A|B) = P(A) \) (or \( P(B|A)=P(B) \)).

1. Calculate total number of cases: \( 7 + 3 + 14 + 2 = 26 \).
2. Calculate \( P(\text{holding soccer practice} | \text{raining}) \):

• Number of cases with raining and practice: \( 7 \).
• Total cases with raining: \( 7 + 3 = 10 \).
• So \( P(\text{practice} | \text{raining})=\frac{7}{10} = 0.70 \).

1. Calculate \( P(\text{holding soccer practice}) \):

• Total cases with practice: \( 7 + 14 = 21 \).
• So \( P(\text{practice})=\frac{21}{26}\approx0.81 \).

1. For independent events, \( P(A|B) \) should equal \( P(A) \). Here, \( 0.70

eq 0.81 \), so the events are dependent.

Now analyze options:

• Option a: Claims independence but values are not close, which is wrong.
• Option b: Correctly states dependence because \( P(\text{practice} | \text{raining}) = 0.70 \) and \( P(\text{practice}) = 0.81 \) (not close), so dependence holds.
• Option c: Comparing \( P(\text{practice}) \) and \( P(\text{raining}) \) is irrelevant for independence (independence is about conditional and marginal probabilities of the same event pair, not comparing two different events' marginal probabilities).
• Option d: The values \( 0.33 \) (incorrect calculation for \( P(\text{raining} | \text{practice}) \): \( \frac{7}{21}\approx0.33 \)) and \( 0.38 \) ( \( \frac{10}{26}\approx0.38 \)) are relatively close, but the reasoning for dependence here is wrong (close values would suggest independence, but also the approach is incorrect as we should check \( P(A|B) \) vs \( P(A) \) or \( P(B|A) \) vs \( P(B) \)).

Answer:

b) They appear to be dependent events because \( P(\text{holding soccer practice} | \text{raining}) = 0.70 \) and \( P(\text{holding soccer practice}) = 0.81 \) and the values are not close to each other.