Abag contains 6 red balls, 4 green balls, and 3 blue balls. if we choose a ball, then another ball without putting the first one back in the bag, what is the probability that the first ball will be green and the second will be red?

Step-by-step explanation:

Given : A bag contains 6 red balls, 4 green balls, and 3 blue balls.

Total balls = 6+4+3=13

Probability of drawing first ball as green :

If the first ball will be green, then the total balls left in bag = 13-1=12 and number of red balls remains the same.

Now, The conditional probability of drawing a red ball given that first ball was green :-

Now,  the probability that the first ball will be green and the second will be red will be :-

Hence, the required probability =

$4034.40

the perimeter of the field is the sum of the lengths of the four sides:

47 yd + 158 yd + 47 yd + 158 yd = 410 yd

the cost of 3 ft (1 yd) of fence is 3×$3.28 = $9.84.

so the cost of jake's wood perimeter fance is

(410 yd) × ($9.84 /yd) = $4034.40