Question

sams closet contains blue and green shirts. he has eight blue shirts, and seven green shirts. five of the blue shirts have stripes, and four of the green shirts have stripes. what is the probability that sam randomly chooses a shirt that is blue or has stripes?
a. $p(\text{b} \cup \text{s}) = 0.8$
b. $p(\text{b} \cup \text{s}) = 0.69$
c. $p(\text{b} \cup \text{s}) = 0.33$
d. $p(\text{b} \cup \text{s}) = 0.27$

Explanation:

Step1: Find total number of shirts

Total blue shirts = 8, total green shirts = 7. So total shirts \( n = 8 + 7 = 15 \).

Step2: Define events and find their counts

Let \( B \) be event of blue shirt, \( S \) be event of striped shirt.
\( n(B) = 8 \), \( n(S) = 5 + 4 = 9 \) (5 blue striped + 4 green striped), \( n(B \cap S) = 5 \) (blue striped shirts).

Step3: Apply principle of inclusion - exclusion

\( P(B \cup S) = P(B) + P(S) - P(B \cap S) \)
\( P(B) = \frac{8}{15} \), \( P(S) = \frac{9}{15} \), \( P(B \cap S) = \frac{5}{15} \)
\( P(B \cup S) = \frac{8}{15} + \frac{9}{15} - \frac{5}{15} = \frac{12}{15} = 0.8 \)? Wait, no, wait: Wait, total shirts are 15? Wait 8 blue +7 green =15. Striped: 5 blue +4 green=9. Blue and striped:5. So \( P(B \cup S)=\frac{n(B)+n(S)-n(B\cap S)}{n}=\frac{8 + 9 - 5}{15}=\frac{12}{15}=0.8 \). Wait but option A is 0.8. Wait maybe I miscalculated? Wait 8+9-5=12, 12/15=0.8. So A is correct? Wait the yellow is on B, but maybe mistake. Wait let's recalculate:

Wait 8 blue, 7 green. Striped: 5 blue, 4 green. So blue or striped: blue shirts (8) + green striped shirts (4) = 12. So 12/15=0.8. So \( P(B \cup S)=0.8 \), which is option A.

Wait maybe the initial yellow highlight was wrong. Let's check again:

Total shirts: 8 +7=15.
Number of blue or striped: blue (8) + green striped (4) =12. 12/15=0.8. So \( P(B \cup S)=0.8 \), so option A.

Answer:

A. \( \boldsymbol{P(\text{B} \cup \text{S}) = 0.8} \)