<span>2/3 probability of drawing a green marble on the second draw.
If the marbles are replaced after each draw, the probability will always be 2/3 of a green marble being drawn. But for this problem, going to assume that the marbles are not replaced after each draw. So we have 2 possible scenarios.
1. The first marble drawn is green
This even happens with probability of 2/3 and leaves you with 7 green marbles and 4 blue marbles. So the probability of picking a green marble again is 7/11.
2. The first marble drawn is blue
This event happens with probability of 1/3 and leaves you with 8 green marbles and 3 blue marbles. So the probability of picking a green marble this time is 8/11
The total probability of picking a green marble on the 2nd pick is the sum of the product of each probability. So
2/3 * 7/11 + 1/3 * 8/11 = 14/33 + 8/33 = 22/33 = 2/3</span>
2
2 ⅓ - ⅓= 2 <em>because</em><em> </em><em>now</em><em> </em><em>you've </em><em>gotten</em><em> </em><em>rid</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>⅓</em><em>.</em>
Answer:
P(A or B) = 1.16
Step-by-step explanation:
Given probability:
Probability of event A = P(A) = 0.46
Probability of event B = P(B) = 0.7
P(A and B) = 0.43
Find:
P(A or B)
Computation:
If A is an incident and B is a separate event, P(A or B) is the possibility of either A, B, or both events occurring.
P(A or B) = P(A) + P(B)
P(A or B) = 0.46 + 0.7
P(A or B) = 1.16
The value of the missing length is 20
Let the missing length be x
From Parallel lines and transversal theorem, we have that
If three or more parallel lines intersect two transversals, then they cut off the transversals proportionally.
Then, we can write that
∴ 
x = 20
Hence, the value of the missing length is 20
Learn more here: brainly.com/question/24258397
