A company sells 14 types of crackers that they label varieties 1 through 14, based on spice level. What is the probability that the purchase results in a selection of a cracker with number less than or equal to 4, or a number greater than 10

Respuesta :

Answer:

[tex]P(x\le 4\ or x> 10) = \frac{4}{7}[/tex] or [tex]P(x\le 4\ or x> 10) = 0.5714[/tex]

Step-by-step explanation:

Given

[tex]x = \{1,2,3,4,5,6,7,8,9,10,11,12,13,14\}[/tex]

Required

Determine [tex]P(x\le 4\ or x> 10)[/tex]

Because the events are independent, the probability can be solved using:

[tex]P(A\ or\ B) = P(A) + P(B)[/tex]

So, we have:

[tex]P(x\le 4\ or x> 10) = P(x \le 4) + P(x > 10)[/tex]

When [tex]x \le 4[/tex], we have: [tex]x = \{1,2,3,4\}[/tex]

So:

[tex]P(x \le 4) = \frac{4}{14}[/tex]

Also:

When [tex]x > 10[/tex], we have: [tex]x = \{11,12,13,14\}[/tex]

So:

[tex]P(x>10) =\frac{4}{14}[/tex]

[tex]P(x\le 4\ or x> 10) = P(x \le 4) + P(x > 10)[/tex] becomes

[tex]P(x\le 4\ or x> 10) = \frac{4}{14} + \frac{4}{14}[/tex]

[tex]P(x\le 4\ or x> 10) = \frac{4+4}{14}[/tex]

[tex]P(x\le 4\ or x> 10) = \frac{8}{14}[/tex]

[tex]P(x\le 4\ or x> 10) = \frac{4}{7}[/tex]

[tex]P(x\le 4\ or x> 10) = 0.5714[/tex]