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]

(a) Compute the probability that a randomly selected peanut M&M is not yellow.
(b) Compute the probability that a randomly selected peanut M&M is orange or yellow.
(c) Compute the probability that three randomly selected peanut M&M's are all red.
(d) If you randomly select two peanut M&M's, compute that probability that neither of them are red.
(e) If you randomly select two peanut M&M's, compute that probability that at least one of them is red.

Answer 1

The probability that a randomly selected peanut M&M is not yellow is 0.85. The probability that a randomly selected peanut M&M is orange or yellow is 0.38. The probability that three randomly selected peanut M&M's are all red is 0.001728.

Step-by-step explanation:

(a) Probability that a randomly selected peanut M&M is not yellow:

To find the probability of an event happening, we subtract the probability of the event not happening from 1. Therefore, the probability that a peanut M&M is not yellow is:

1 - 0.15 = 0.85

(b) Probability that a randomly selected peanut M&M is orange or yellow:

To find the probability of the union of two events happening, we add their individual probabilities. Therefore, the probability that a peanut M&M is orange or yellow is:

0.23 + 0.15 = 0.38

(c) Probability that three randomly selected peanut M&M's are all red:

Since the selection is done with replacement, the probability of selecting a red M&M in one draw is 0.12. Therefore, the probability that three randomly selected peanut M&M's are all red is:

0.12 * 0.12 * 0.12 = 0.001728

(d) Probability that neither of the two selected peanut M&M's are red:

The probability of selecting a non-red M&M in one draw is 1 - 0.12 = 0.88. Therefore, the probability that neither of the two selected peanut M&M's are red is:

0.88 * 0.88 = 0.7744

(e) Probability that at least one of the two selected peanut M&M's is red:

To find the probability of at least one event happening, we subtract the probability of none of the events happening from 1. Therefore, the probability that at least one of the two selected peanut M&M's is red is:

1 - 0.7744 = 0.2256

Answer 2

(a)0.85

(b)0.38

(c)0.002

(d)0.986

(e)0.12

Step-by-step explanation:

Brown = 12%,

Yellow = 15%

Red = 12%

Blue = 23%

Orange= 23%

Green = 15%

(a) Probability that a randomly selected peanut M&M is not yellow.

P(Yellow) = 15/100 =0.15

Therefore: P(Not Yellow)

=1-0.15 =0.85

(b) Probability that a randomly selected peanut M&M is orange or yellow.

P(Orange OR Yellow)

= P(Orange)+P(Yellow)

=23/100 + 15/100

=38/100 =0.38

(c) Probability that three randomly selected peanut M&M's are all red.

P(Red and Red and Red) = 12/100 X 12/100 X 12/100 =0.001728 = 0.002

(d) Probability that neither of them are red.

P(neither of them are red)= 1- P(both are red) = 1-(12/100 X 12/100)

= 1-0.0144 = 0.9856 = 0.986

(e)Probability that at least one of them is red.

P(RB or RY or RR or Rb or RO or RG)

=(0.12 X 0.12)+(0.12 X 0.15)+(0.12 X 0.12)+(0.12 X 0.23)+(0.12 X 0.23)+ (0.12 X 0.15)

= 0.12