contestada

3 rooms are occupied by disease x, 4 by disease y, 2 rooms by disease z and 1 by disease a. What is the probability that a random room does not contain disease y?

Respuesta :

ronhagrid310

Answer:

6/10

Step-by-step explanation:

[P(x) + P(z) + P(a)]/P(all) = (3+2+1)/(3+4+2+1) = 6/10

arianatingz

Answer:

6/10

Step-by-step explanation:

dhdhdhcebwbshxhdnsncmfnnsjcndnendn

Otras preguntas