Question

A bag of 11 marbles contains 6 marbles with red on them, 6 with blue on them, 6 with green on them, and 2 with red and green on them. What is the probability that a randomly chosen marble has either green or red on it? Note that these events are not mutually exclusive. Express your answer as a fraction in lowest terms or a decimal rounded to the nearest millionth.

Answer 1

To calculate the probability of drawing a marble that is either red or green from the bag, apply the Addition Rule for probability, accounting for double-counting overlapping events. The calculation results in a probability of 10/11 after simplification.

Step-by-step explanation:

To calculate the probability that a randomly chosen marble has either green or red on it, we can use the Addition Rule for probabilities because the events are not mutually exclusive, meaning a marble can have both colors. The probability of drawing a green marble (P(Green)) plus the probability of drawing a red marble (P(Red)) minus the probability of drawing a marble that is both red and green (P(Red and Green)) will give us the probability of drawing a marble that is either red or green (P(Red or Green)).

Firstly, we need to correct the total number of marbles, as the provided numbers initially imply there are more marbles than there actually are. To find the correct total, we can assume that there are 2 marbles that are both red and green, which are being counted twice in the total for red and green marbles. So, we subtract this overlap from the initial total (6 red + 6 green - 2 red and green) to avoid double-counting those marbles. Therefore, the total number of unique marbles is 11.

Here are the steps to find P(Red or Green):

1. Calculate P(Red) = Number of red marbles / Total number of marbles = 6/11
2. Calculate P(Green) = Number of green marbles / Total number of marbles = 6/11
3. Calculate P(Red and Green) = Number of red and green marbles / Total number of marbles = 2/11
4. Apply the Addition Rule: P(Red or Green) = P(Red) + P(Green) - P(Red and Green) = 6/11 + 6/11 - 2/11
5. Simplify to find the answer: P(Red or Green) = 10/11

Answer 2

`\begin{gathered} P(A\cup B)=P(A)+P(B)-P(A\cap B) \\ P(A)=Probability\text{ marble has gre}en \\ P(B)=Probability\text{ marble has red} \\ P(A\cap B)=Probability\text{ marble has gre}en\text{ and red} \\ P(A)=(6)/(11) \\ P(B)=(6)/(11) \\ P(A\cap B)=(2)/(11) \\ P(A\cup B)=(6)/(11)+(6)/(11)-(2)/(11) \\ P(A\cup B)=(10)/(11) \\ \text{The probability of chosen a marble that has either gr}een\text{ } \\ \text{or red on it is }(10)/(11) \end{gathered}`