Question

according to masterfoods, the company that manufactures m&ms, 12% of peanut m&ms are brown, 15% are yellow, 12% are red, 23% are blue, 23% are orange and 15% are green. (round your answers to 4 decimal places where possible)
a. compute the probability that a randomly selected peanut m&m is not green.
b. compute the probability that a randomly selected peanut m&m is orange or brown.
c. compute the probability that two randomly selected peanut m&ms are both brown.
d. if you randomly select six peanut m&ms, compute that probability that none of them are yellow.
e. if you randomly select six peanut m&ms, compute that probability that at least one of them is yellow.

Explanation:

Step1: Find probability of not green

The probability of an M&M being green is $P(G)=0.15$. The probability of not being green is $P(\text{not }G)=1 - P(G)$.
$P(\text{not }G)=1 - 0.15=0.8500$

Step2: Find probability of orange or brown

The probability of an M&M being orange is $P(O)=0.23$ and being brown is $P(B)=0.12$. Since they are mutually - exclusive events, $P(O\cup B)=P(O)+P(B)$.
$P(O\cup B)=0.23 + 0.12=0.3500$

Step3: Find probability of two brown M&Ms

The probability of one M&M being brown is $P(B)=0.12$. Since the selections are independent, the probability of two being brown is $P(B_1\cap B_2)=P(B)\times P(B)$.
$P(B_1\cap B_2)=0.12\times0.12 = 0.0144$

Step4: Find probability of no yellow in six selections

The probability of an M&M not being yellow is $P(\text{not }Y)=1 - 0.15 = 0.85$. Since the selections are independent, the probability of none of six being yellow is $(P(\text{not }Y))^6$.
$(P(\text{not }Y))^6=0.85^6\approx0.3771$

Step5: Find probability of at least one yellow in six selections

The probability of at least one yellow is the complement of the event of no yellow. Let $A$ be the event of at least one yellow. Then $P(A)=1 - P(\text{no yellow})$.
$P(A)=1 - 0.85^6\approx1 - 0.3771 = 0.6229$

Answer:

a. $0.8500$
b. $0.3500$
c. $0.0144$
d. $0.3771$
e. $0.6229$