Respuesta :
Total number of computers in sample = 72 + 112 + 711 = 895.
P(LQ) = 72/895 = 0.08
P(MQ) = 112/895 = 0.125
P(HQ) = 711/895 = 0.794
P(LQ) = 72/895 = 0.08
P(MQ) = 112/895 = 0.125
P(HQ) = 711/895 = 0.794
Answer:
[tex]P(L) = 0.0804[/tex]
[tex]P(M) = 0.1251[/tex]
[tex]P(H) = 0.7945[/tex]
Step-by-step explanation:
A probability is the number of desired outcomes desired by the number of total outcomes.
I am going to say that we have these following events
L is the event that a computer is low quality.
There are 72+112+711 = 895 computers. Of those, 72 are low quality. So
[tex]P(L) = \frac{72}{895} = 0.0804[/tex]
M is the event that a computer is medium quality.
Of the 895 computers, 112 of them are medium quality. So
[tex]P(M) = \frac{112}{895} = 0.1251[/tex]
H is the event that a computer is medium quality.
Of the 895 computers, 711 of them are medium quality. So
[tex]P(H) = \frac{711}{895} = 0.7945[/tex]