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A bag of 11 marbles contains 7 marbles with red on them, 4 with blue on them, 6 with green on them and 6 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.

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Answer:

7/11

Step-by-step explanation:

Given;

Total number of marbles = 11

Number of marbles with red on them = 7

Number of marbles with blue on them = 4

Number of marbles with green on them = 6

Number of marbles with red and green on them = 6

We'll use the below formula to determine the probability that a randomly chosen marble has either green or red on it;


P(G\cup R)=P(G)+P(R)-P(G\cap R)

where;


\begin{gathered} P(G\cup R)=\text{ the probabilty of choosing a marble with green or red on it = ?} \\ P(G)=\text{ the probability of choosing marbles with gre}en\text{ on them }=(6)/(11) \\ P(R)=\text{the probability of choosing marbles with red on them}=(7)/(11) \\ P(G\cap R)=\text{the probability of choosing marbles with gre}en\text{ and red on them }=(6)/(11) \end{gathered}

Let's go ahead and substitute the above values into the formula and evaluate;


\begin{gathered} P(G\cup R)=(6)/(11)+(7)/(11)-(6)/(11) \\ P(G\cup R)=(7)/(11) \end{gathered}

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