Question

according to masterfoods, the company 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 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 green.
c. compute the probability that two randomly selected peanut m&m’s are both blue.
d. if you randomly select six peanut m&m’s, compute that probability that none of them are brown.
e. if you randomly select six peanut m&m’s, compute that probability that at least one of them is brown.

Explanation:

Step1: Recall probability of complementary event

The probability of an event $A$ not occurring, $P(\text{not }A)=1 - P(A)$. For part a, let $A$ be the event that an M&M is green. Given $P(A) = 0.15$, then $P(\text{not green})=1 - 0.15$.
$1-0.15 = 0.8500$

Step2: Use addition - rule for mutually - exclusive events

For part b, if $A$ is the event that an M&M is orange and $B$ is the event that an M&M is green, and $A$ and $B$ are mutually - exclusive (an M&M can't be both orange and green at the same time), then $P(A\cup B)=P(A)+P(B)$. Given $P(A) = 0.23$ and $P(B)=0.15$, so $P(A\cup B)=0.23 + 0.15$.
$0.23+0.15 = 0.3800$

Step3: Use multiplication rule for independent events

For part c, if the selection of the first and second M&M are independent events, and the probability that a single M&M is blue is $p = 0.23$, then the probability that two M&Ms are blue is $P(\text{both blue})=0.23\times0.23$.
$0.23\times0.23=0.0529$

Step4: Use multiplication rule for independent events

For part d, the probability that a single M&M is not brown is $1 - 0.12=0.88$. Since the selections of the six M&Ms are independent events, the probability that none of the six M&Ms are brown is $(0.88)^6$.
$(0.88)^6=0.4644$

Step5: Use the complement rule

For part e, let $A$ be the event that at least one of the six M&Ms is brown. The complement of $A$ is the event that none of the six M&Ms are brown. We found in part d that the probability that none of the six M&Ms are brown is $(0.88)^6\approx0.4644$. Then $P(A)=1-(0.88)^6$.
$1 - 0.4644 = 0.5356$

Answer:

a. $0.8500$
b. $0.3800$
c. $0.0529$
d. $0.4644$
e. $0.5356$