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]

Answer 1

Answer:

(a) P=0.850 (b) P=0.460 (c) P=0.053 (d) P=0.457 (e) P=0.543

Step-by-step explanation:

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

$P(X\neq G) = 1-P(X=G) = 1-0.15=0.850$

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

$P(X=B or X=O) = P(X=B)+P(X=O) = 0.23+0.23 = 0.460$

 

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

$P(X1=O\& X2=O) = P(X=O)^2 = 0.23^2 = 0.053$

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

$P(X1 \neq O \& X2\neq O \& X3\neq O) = (1-P(X=O)^3 = (1-0.23)^3 = 0.77^3 = 0.457$

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

We have to compute the probability that only one peanut is orange, plus the probaibility that only two peanuts are orange plus the probability that the three peanuts are orange

$P(X1=O or X2=O or X3=O) = 3*P(X=O)*P(X≠O)^2 + 3*P(X=O)^2* P(X≠O)+P(X=O)^3\\\\P(X1=O or X2=O or X3=O) =3*0.23*0.77^2+3*0.23^2*0.77+0.23^3\\\\P(X1=O or X2=O or X3=O) = 0.409+0.122+0.012=0.543$