A snack-size bag of M&Ms candies is opened. Inside, there are 12 red candies, 12 blue, 7 green, 13 brown, 3 orange, and 10 yellow. Three candies are pulled from the bag in succession, without replacement. What is the probability that the first two candies drawn are orange and the third is green?

Step-by-step explanation:

GIVEN: A snack-size bag of M&Ms candies is opened. Inside, there are red candies, blue, green, brown, orange, and yellow. Three candies are pulled from the bag in succession, without replacement.

TO FIND: What is the probability that the first two candies drawn are orange and the third is green.

SOLUTION:

Total  candies in the bag

Probability that first ball is orange,

Probability that second ball is orange,

Probability that third ball is green,

Now, probability that first two balls are orange and third is green is

Hence,  probability that first two balls are orange and third is green is

(1,4) is on graph of f(x)

f(1) = 4

so 1/4 f(x)

1/4 f(1)

1/4. ×4

1

so point for function 1/4. f(x) is (1,1)

idk this answer sorry cant on this one.

step-by-step explanation: