Hey buddy I am here to help!
yellow marbles = 4
blue marbles = 2
red marbles = 3
total marbles = 9
1 random marbles taken off probability left = 8/9
1 random marble taken off probability = 1/9
marble if replaced = + 1
if one marble if replaced then = 8/9 + 1/9 = 9/9 that means total marbles r 9
yellow marbles drawn off = 2
so... 4 - 2 = 2 marbles left
red marbles taken off = 1
so... 3 -1 = 2 marbles left
probability/answer = 3/9 = 1/3
Hope it helps!
Plz mark my answer brainliest
Answer:
The product of
36√cis(π/8) and 25√cis(7π/6)
is
(225√2)√[√(2 + √2) + i√(2 - √2)][√(3(-1 + i))]
Step-by-step explanation:
First note that
cis(π/8) = cos(π/8) + isin(π/8)
cis(7π/6) = cos(7π/6) + isin(7π/6)
cos(π/8) = (1/2)√(2 + √2)
sin(π/8) = (1/2)√(2 - √2)
36√cis(π/8) = (36/√2)√[√(2 + √2) + i√(2 - √2)]
cos(7π/6) = -(1/2)√3
sin(7π/6) = (1/2)√3
25√cis(7π/6) = (25/2)√3(-1 + i)
The product,
36√cis(π/8) × 25√cis(7π/6)
= (36/√2)√[√(2 + √2) + i√(2 - √2)] × (25/2)√3(-1 + i)
= (225√2)√[√(2 + √2) + i√(2 - √2)][√(3(-1 + i))]
Answer:
The probability is 0.3576
Step-by-step explanation:
The probability for the ball to fall into the green ball in one roll is 2/1919+2 = 2/40 = 1/20. The probability for the ball to roll into other color is, therefore, 19/20.
For 25 rolls, the probability for the ball to never fall into the green color is obteined by powering 19/20 25 times, hence it is 19/20^25 = 0.2773
To obtain the probability of the ball to fall once into the green color, we need to multiply 1/20 by 19/20 powered 24 times, and then multiply by 25 (this corresponds on the total possible positions for the green roll). The result is 1/20* (19/20)^24 *25 = 0.3649
The exercise is asking us the probability for the ball to fall into the green color at least twice. We can calculate it by substracting from 1 the probability of the complementary event: the event in which the ball falls only once or 0 times. That probability is obtained from summing the disjoint events: the probability for the ball falling once and the probability of the ball never falling. We alredy computed those probabilities.
As a result. The probability that the ball falls into the green slot at least twice is 1- 0.2773-0.3629 = 0.3576