AYUDAAA porfavo primera urna 6 bolas verdes y 3 rojas segunda urna 3 verdes, 3 blancas y 3 rojas tercera urna 6 verdes, 1 blanca y 2 rojas a) si sacan una bola de la primera urna que color es más probable de que se obtenga b) calculen las probabilidades de sacar una bola verde, una roja y una blanca de cada urna c) si quieren tomar una bola verde que urna elegirian d) supongan que sacan una bola verde de la tercera urna y no la regresan, lo que se conoce como extraccion sin reemplazo ¿que probabilidad tiene cada bola de salir en la siguiente extraccion?

We have the following information:

first urn: 6 green balls and 3 red ones

total: 6 + 3 = 9

second urn: 3 green, 3 white and 3 red

total: 3 + 3 + 3 = 9

third urn: 6 green, 1 white and 2 red

total: 6 + 1 + 2 = 9

a) A green ball is more likely to be obtained, since there are more green balls than red balls, which makes the probability higher.

b) probability of drawing a green, red and white ball.

first urn:

green = 6/9 = 66.66%

red = 3/9 = 33.33%

white = 0/9 = 0%

second urn:

green = 3/9 = 33.33%

red = 3/9 = 33.33%

white = 3/9 = 33.33%

third urn:

green = 6/9 = 66.66%

red = 2/9 = 22.22%

white = 1/9 = 11.11%

c) it would be chosen where the probability of drawing green would be the highest, which means that it would be possible both in the first and in the third ballot box, the probability is equal 66.66%

d) without a green ball, the third ballot box would look like this:

5 green balls, 2 red balls and 1 white ball, with a total of 8.

The probability of drawing would be:

green = 5/8 = 62.5%

red = 2/8 = 25%

white = 1/8 = 12.5%

168

step-by-step explanation:

i am pretty sure that the answer is 168