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%
I believe the answer is D.
Answer:
108 1/8
Step-by-step explanation:
1/4=0.25, 1/5=0.2
=>26+0.2=26.2
=>26.2/0.25=108.8
Answer:
Step-by-step explanation:
this has bean simplifyed dow to
= -1/6
I think it’s (-5,4) i hope this helps!
