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

the median here is 25. your lowest value is 14 and highest value is 48. to find the third quartile, find the medium number in between 25 and 48 which would be 42 i believe.

1gm = 1000mg so 0.1gm = 1000mg

first, convert the weight to kg

160lb * 1kg/2.2lb = 72.7kg

next, the dose

72.7kg * 100mg/kg = 7270mg