Faculty positions in a school district are classified as either teacher or administrator based on primary duties and responsibilities. The table below shows information on the number of national board-certified and not board-certified teachers and administrators

Board-certified Not board-certified
Administrator 17 19
Teacher 13 12

Determine the probability, P(board-certified or teacher), P(board-certified or teacher), that a randomly chosen faculty member is either board-certified or a teacher. Please enter your answer as a decimal, precise to at least four decimal places.

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amna04352 amna04352 · Answer 1 · 2020-03-19T14:11:29+01:00

Answer:

42/61

Step-by-step explanation:

(17+13+12)/(17+13+12+19)

= 42/61 or 0.6885

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MrRoyal MrRoyal · Answer 2 · 2021-09-17T18:38:51+02:00

Probabilities are used to determine the chance of an event.

The probability that a randomly chosen faculty member is either board-certified or a teacher is 0.6885

Let:

[tex]B \to[/tex] Board-certified

[tex]T \to[/tex] Teacher

[tex]NB \to[/tex] Not board certified

[tex]NT \to[/tex] Not teacher

From the table, we have:

[tex]B = 17 + 13 = 30[/tex]

[tex]T = 13 + 12 = 25[/tex]

[tex]B\ n\ T = 13[/tex]

[tex]Total = 17 + 19 + 13 + 12[/tex]

[tex]Total = 61[/tex]

The required probability is represented as:

P(board-certified or teacher) [tex]\to[/tex] P(B or T)

And it is calculated using the following formula:

[tex]P(B\ or\ T) =P(B) + P(T) -P(B\ and\ T)[/tex]

So, we have:

[tex]P(B\ or\ T) = \frac{B + T - B\ and\ T}{Total}[/tex]

[tex]P(B\ or\ T) = \frac{30 + 25 - 13}{61}[/tex]

[tex]P(B\ or\ T) = \frac{42}{61}[/tex]

[tex]P(B\ or\ T) = 0.6885[/tex]

Hence, the required probability is 0.6885