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 three decimal places, for example: 0.123 compute the probability that a randomly selected peanut m&m is not red. compute the probability that a randomly selected peanut m&m is red or green. compute the probability that three randomly selected peanut m&ms are all green. if you randomly select four peanut m&ms, compute that probability that none of them are green. if you randomly select four peanut m&ms, compute that probability that at least one of them is green.

Explanation:

Step1: Probability that a peanut M&M is not red

We know that the probability of an event \(A\) not occurring is \(P(\text{not }A)=1 - P(A)\).
The probability that a peanut M&M is red \(P(\text{red}) = 0.12\).
So \(P(\text{not red})=1 - 0.12\)

$$P(\text{not red})=0.88$$

Step2: Probability that a peanut M&M is red or green

For two mutually - exclusive events \(A\) and \(B\) (if \(A\cap B=\varnothing\)), \(P(A\cup B)=P(A)+P(B)\)
The probability that a peanut M&M is red \(P(\text{red}) = 0.12\) and the probability that a peanut M&M is green \(P(\text{green})=0.15\)
So \(P(\text{red or green})=0.12 + 0.15\)

$$P(\text{red or green})=0.27$$

Step3: Probability that three randomly - selected peanut M&M's are all green

If the selections are independent, and for independent events \(A\), \(B\), \(C\), \(P(A\cap B\cap C)=P(A)\times P(B)\times P(C)\)
The probability that a peanut M&M is green \(P(\text{green}) = 0.15\)
So \(P(\text{all three green})=0.15\times0.15\times0.15\)

$$P(\text{all three green})=0.15^{3}=0.003375\approx0.003$$

Step4: Probability that none of four randomly - selected peanut M&M's are green

The probability that a peanut M&M is not green \(P(\text{not green})=1 - 0.15=0.85\)
For four independent selections, \(P(\text{none green})=(0.85)^{4}\)

$$P(\text{none green})=0.85\times0.85\times0.85\times0.85 = 0.52200625\approx0.522$$

Step5: Probability that at least one of four randomly - selected peanut M&M's is green

Let \(A\) be the event that at least one is green. Then \(A\) is the complement of the event that none is green.
We know \(P(A)=1 - P(\text{none green})\)
Since \(P(\text{none green})=(0.85)^{4}\approx0.522\)
So \(P(\text{at least one green})=1-(0.85)^{4}\)

$$P(\text{at least one green})= 1 - 0.522=0.478$$

Answer:

\(0.880\)
\(0.270\)
\(0.003\)
\(0.522\)
\(0.478\)