Mark is going to an awards dinner and wants to dress appropriately. He is running behind schedule and asks his little brother to randomly select an outfit for him. Mark has one blue dress shirt, one white dress shirt, one black dress shirt, one pair of black slacks, one pair of grey slacks, and one red tie. All six of his possible outfits are listed below. Let A AA be the event that Mark's little brother selects an outfit with black slacks and B BB be the event that he selects an outfit with a blue shirt. What is P ( A or B ) P(A or B)P, left parenthesis, A, start text, space, o, r, space, end text, B, right parenthesis, the probability that Mark's little brother selects an outfit with black slacks or an outfit with a blue shirt? Shirt Slacks Tie Outfit 1 11 Blue Black Red Outfit 2 22 Blue Grey Red Outfit 3 33 White Black Red Outfit 4 44 White Grey Red Outfit 5 55 Black Black Red Outfit 6 66 Black Grey Red

Answer:

The answer is "[tex]\bold{\frac{2}{3}}[/tex]".

Step-by-step explanation:

Pick a shirt first: 3 choices

Pick slacks next: 2 choices

Then choose a tie: 1 choice

Total: [tex]3 \times 2 \times 1 = 6[/tex] outfits.

These will be similarly divided between the two slacks, so 3 black slack outfits of [tex]6 = \frac{1}{2}[/tex].  Similarly, the 6 styles are also divided between the option of 3 shirts. 2 blue shirt outfits [tex]6 = \frac{1}{3}[/tex]

To recognize that over outfits are a mix of a blue shirt and black slack, it would be twice many more to add [tex]\frac{1}{2} \ and \ \frac{1}{3}[/tex]. Add [tex]\frac{1}{2} \ and \ \frac{1}{3}[/tex], and remove the chance for both the black slack shirt and outfit and are one of the six outfits:

[tex]\to \frac{1}{2} + \frac{1}{3} -\frac{1}{6} =\frac{3+2-1}{6} =\frac{4}{6} = \frac{2}{3}[/tex]

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