College

Sain's 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 Sain randomly chooses a shirt that is blue or has stripes?

A. [tex]P(B \cup S) = 0.33[/tex]
B. [tex]P(B \cup S) = 0.8[/tex]
C. [tex]P(B \cup S) = 0.27[/tex]
D. [tex]P(B \cup S) = 0.69[/tex]

Answer :

We are given the following information:

- Total number of shirts is
[tex]$$8 \text{ (blue)} + 7 \text{ (green)} = 15.$$[/tex]
- Number of blue shirts is 8.
- The total number of shirts with stripes is
[tex]$$5 \text{ (blue striped)} + 4 \text{ (green striped)} = 9.$$[/tex]
- Notice that the blue striped shirts (5) have been counted twice if we simply add blue and striped shirts. To avoid double counting, we use the inclusion‐exclusion principle.

The number of shirts that are either blue or have stripes is:
[tex]$$
\text{Blue or Striped} = (\text{Blue Shirts}) + (\text{Striped Shirts}) - (\text{Blue Striped Shirts})
$$[/tex]

Substitute the numbers:
[tex]$$
\text{Blue or Striped} = 8 + 9 - 5 = 12.
$$[/tex]

Now, the probability that a randomly chosen shirt is blue or has stripes is:
[tex]$$
P(B \cup S) = \frac{\text{Number of Blue or Striped Shirts}}{\text{Total Number of Shirts}} = \frac{12}{15} = 0.8.
$$[/tex]

Thus, the correct answer is:

[tex]$$
P(B \cup S)=0.8.
$$[/tex]