Given that a bag of M & Ms has 21 brown candies, 15 orange candies, 10 yellow candies, 14 red candies, and 5 green candies, find the following probabilities. Enter all answers as reduced fraction. Do not use any spaces in your answers. What is the probability of choosing a green M & M, eating it, and then choosing a brown M & M?
P( Answer
B | G
)= Answer

What is the probability of choosing a red M & M on the second try given that the first M & M was red and it was eaten?

Step-by-step explanation:

These both are independent events. Neither is really a conditional probability. It's like asking "What the chances of getting a heads given that you got a tails in the previous through?" The answer is 1/2 if it is a fair coin.

So the total number of M and Ms is

21+ 15 + 10 + 14 + 5 = 65

5/65 * 14/64 = 0.0168  Be careful you you put that in your calculator.

Part B is a bit harder.

14/65 * 13/64 = 0.04375

step-by-step explanation:

5×5×5+2(8)÷2(2)+3(3)

15+16÷4+9

31÷13

do 31÷13 and u will get ur answer

need more info

step-by-step explanation:

what do you mean by "representation"? graph? table? etc.