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Suppose that a box contains 45 light bulbs, of which 36 are good and the other 9 are defective. Consider randomly selecting three bulbs without replacement. Let E denote the event that the first bulb selected is good, F be the event that the second bulb is good, and G represent the event that the third bulb selected is good.

(a)

What is

P(E)?


(b)

What is

P(F|E)?

(Round your answer to three decimal places.)


(c)

What is

P(G|E ∩ F)?

(Round your answer to three decimal places.)


(d)

What is the probability that all three selected bulbs are good? (Round your answer to three decimal places.)

Respuesta :

Answer:

Step-by-step explanation:

Given that a box contains 45 light bulbs, of which 36 are good and the other 9 are defective.

Probability for any one to be defective as of now is [tex]\frac{36}{45} =0.80[/tex]

a) [tex]P(E) = 0.80[/tex]

b) After I draw we have 35 good ones and 9 defective

P(F/E) =[tex]\frac{35}{44} \\=0.796[/tex]

c) P(G/EF)

This is the probability for iii bulb to be good given I and II are goog.

Once I and ii are good we have 43 bulbs with 34 good ones

[tex]P(G/EF) =\frac{34}{43} \\=0.7907[/tex]

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