Question

According to Masterfoods, the company that manufactures M&M’s, 12% of peanut M&M’s are brown, 15% are yellow, 12% are red, 23% are blue, 23% are orange and 15% are green. [Round your answers to three decimal places, for example: 0.123]

Compute the probability that a randomly selected peanut M&M is not green.




Compute the probability that a randomly selected peanut M&M is blue or orange.




Compute the probability that two randomly selected peanut M&M’s are both orange.




If you randomly select three peanut M&M’s, compute that probability that none of them are orange.




If you randomly select three peanut M&M’s, compute that probability that at least one of them is orange.

Answer 1

you can just as easily consider the % values as integers.

Then just count the ones you need, out of 100

Answer 2

Examples

P(not green) = 1 - .15 = ?

Either-or probabilities are found by adding the individual probabilities.

P(blue or orange) = P(blue) + P(orange)

If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.

P(both orange) = P(orange)^2

P(none orange) = (1-.23)^3

"at least one" = 1, 2 or 3 are orange. Use methods above.

Online "^" is used to indicate an exponent, e.g., x^2 = x squared.

Answer 3

I need answers

Answer 4

Compute the probability that two randomly selected peanut M&M’s are both orange.
Answer:
.23 x .23= .0529

Answer 5

If you randomly select three peanut M&M’s, compute that probability that none of them are orange.
Answer:
1-.23=.77
.77 x .77 x .77 = .4565