according to masterfoods, the company that manufactures m&ms, 12% of peanut m&ms are brown, 15% are yellow…

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.

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.

Answer

Answer:

  1. (0.880)
  2. (0.270)
  3. (0.003)
  4. (0.522)
  5. (0.478)

Explanation:

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

The probability that a peanut M&M is red (P(R)=0.12). Using the complement rule (P(\text{not }A)=1 - P(A)), we have (P(\text{not red})=1 - 0.12=0.88)

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

The probability that a peanut M&M is red (P(R) = 0.12) and the probability that a peanut M&M is green (P(G)=0.15). Using the addition rule for mutually - exclusive events (P(A\cup B)=P(A)+P(B)) (since a peanut M&M can't be red and green at the same time), we get (P(R\cup G)=0.12 + 0.15=0.27)

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

Assume independence. If the probability that one peanut M&M is green (p = 0.15), then for three independent events, using the multiplication rule for independent events (P(A\cap B\cap C)=P(A)\times P(B)\times P(C)), we have (P(\text{all three green})=(0.15)^3=0.15\times0.15\times0.15 = 0.003375\approx0.003)

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

The probability that a peanut M&M is not green (q=1 - 0.15 = 0.85). For four independent events, using the multiplication rule for independent events (P(\text{none green})=(0.85)^4=0.85\times0.85\times0.85\times0.85=0.52200625\approx0.522)

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

Using the complement rule. Let (A) be the event that at least one is green. The complement of (A) is the event that none are green. We know from Step4 that (P(\text{none green})=(0.85)^4). So (P(\text{at least one green})=1-(0.85)^4=1 - 0.52200625 = 0.47799375\approx0.478)