Box A contains 5green and 7 red balls. Box B contains 3green, 3 red and 6 yellow balls. A box is sleeted at random and a ball is drawn at random from it. What is the probability that the drawn ball is green?

Step-by-step explanation:

Given

the sample space for box A

green balls = 5

red balls= 7

sample size= 5+7= 12

the sample space for box B

green balls = 3

red balls= 3

yellow balls= 6

sample size= 3+3+6= 12

The probability of drawing a green ball from box A= 5/12

The probability of drawing a green ball from box B= 3/12= 1/4

Therefore the probability of picking a green ball from either of the boxes at random is =

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116

answer: the resulting cross section is a rectangle. solution: a plane parallel to the base of a triangular prism will intersect a cross section that is the same shape as its bases. so the cross section is a triangle.

step-by-step explanation: