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Answer:
[tex]P(Green\ candy\ or\ Blue\ candy) = 0.16[/tex]
Yes, they are mutually exclusive
Step-by-step explanation:
Given
Refer to the table in the question
Solving (a): P(green candy or blue candy)
This is calculated as:
[tex]P(Green\ candy\ or\ Blue\ candy) = P(Green) + P(Blue)[/tex]
From the question:
[tex]P(Blue) = 10\%[/tex]
[tex]P(Green) = 6\%[/tex]
So:
[tex]P(Green\ candy\ or\ Blue\ candy) = P(Green) + P(Blue)[/tex]
[tex]P(Green\ candy\ or\ Blue\ candy) = 6\% + 10\%[/tex]
[tex]P(Green\ candy\ or\ Blue\ candy) = 16\%[/tex]
Convert to decimal
[tex]P(Green\ candy\ or\ Blue\ candy) = 0.16[/tex]
Solving (b): Are they mutually exclusive?
Multiple events A and B are mutually exclusive if;
[tex]P(A\ or\ B) = P(A) + P(B)[/tex]
In (a) above:
[tex]P(Green\ candy\ or\ Blue\ candy) = P(Green) + P(Blue)[/tex]
and this is so every other possible colour of candy
e.g.
[tex]P(Red\ or\ Yellow) = P(Red) + P(Yellow)[/tex]
Hence:
The events are mutually exclusive