Question

Anthony has two bags of marbles. the first bag contains 12 red marbles, and 8 green marbles. The second bad contains 6 red marbles, 4 blue marbles, and 8 green marbles. Anthony will randomly select 1 marble from each bag. What is the probability that he will select a red marble from each bag

Answer 1

Answer: for the first bag its a 60% chance for the second bag its a 33.33% chance

Step-by-step explanation: if you add the numbers from bag one together, you get 20. and 12 is 60% of twenty.

for bag two if you add the numbers together you get 18. and 6 is 33.33% of 18.

hope this helps

Answer 2

First bag:

Red = 12

Green = 8

Second bag:

Red = 6

Blue = 4

Green = 8

The probability to select a red marble from each bag can be calculated using:

`P(\text{Red})=\frac{\text{ Number of reds}}{\text{ Total}}`

Now, for the first bag:

`P_1(\text{Red})=(12)/(20)=(3)/(5)`

Now, for the second bag:

`P_2(\text{Red})=(6)/(18)=(1)/(3)`

Both selections are independent events, so the overall probability to select a red marble from each bag is just the product of P₁ and P₂:

`\begin{gathered} P(\text{Red})=P_1(\text{Red})\cdot P_2(\text{Red})=(3)/(5)\cdot(1)/(3) \\ P(\text{Red})=(1)/(5) \end{gathered}`