Question

en una canasta se tienen 10 bolas cafes, 5 bolas azules y 15 verdes. Si se saca una al azar, ¿cual es la probabilidad de que esta no sea azul? ¿cual es probabilidad de que sea verde?

Answer 1

The probability that the selected ball is not blue is
`(5)/(6)`.

The probability that the selected ball is green is
`(1)/(2)`.

Explanation:

The question is:

There are 10 brown balls, 5 blue balls and 15 green balls in a basket. If one is drawn at random, what is the probability that it is not blue? What is the probability that it is green?

Solution:

The probability of an event E is the ratio of the favorable number of outcomes to the total number of outcomes.

`P(E)=(n(E))/(N)`

The probability of the given event not taking place is known as the complement of that event.

Complement of the event E is,

1 – P (E)

The number of different color balls are as follows:

Brown = n (Br) = 10

Blue = n (Bu) = 5

Green = n (G) = 15

Total = N = 30

Compute the probability of selecting a blue ball as follows:

`P(\text{Bu})=\frac{n(\text{Bu})}{N}=(5)/(30)=(1)/(6)`

Compute the probability of not selecting a blue ball as follows:

`P(\text{Not Bu})=1-P(\text{Bu})`

`=1-(1)/(6)\\\\=(6-1)/(6)\\\\=(5)/(6)`

Thus, the probability that the selected ball is not blue is
`(5)/(6)`.

Compute the probability of selecting a green ball as follows:

`P(\text{G})=\frac{n(\text{G})}{N}=(15)/(30)=(1)/(2)`

Thus, the probability that the selected ball is green is
`(1)/(2)`.