A box contains 21 marbles of the same size. 7 of the marbles are white, 4 are black, 10 are green. If two of the marbles are randomly selected from the box, what is the probability that the first marble is white and the second is green, given that the first marble is not replaced before the second marble is selected

7/21 x 10/20

1/3 x 1/2 = 1/6

Step-by-step explanation:

Probability calculates the likelihood of an event occurring. The likelihood of the event occurring lies between 0 and 1. It is zero if the event does not occur and 1 if the event occurs.

For example, the probability that it would rain on Friday is between o and 1. If it rains, a value of one is attached to the event. If it doesn't a value of zero is attached to the event.

probability of picking a white marble first = total number of white marbles / total number of marbles = 7/21

probability of picking a green marble second = total number of green marbles / total number of marbles = 10/20

7/21 x 10/20

1/3 x 1/2 = 1/6

93.8

step-by-step explanation:

lets say, first number = x

second number = 3x+7

total = 49

(x)+(3x+7)= 49

4x+7 = 49

4x = 42

x = 42/4

x= 10.5

x= 10.5

3x+7 = 3(10.5)+7 = 93.8

hope this

the answer is 45 i just did it on oddesy ware