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. you randomly select five peanut m&ms from an extra - large bag of the candies. (round all probabilities below to four decimal places: your answer should look like 0.1234, not 0.1234444 or 12.34%.) compute the probability that exactly four of the five m&ms are red: compute the probability that three or four of the five m&ms are red: compute the probability that at most four of the five m&ms are red: compute the probability that at least four of the five m&ms are red: if you repeatedly select random samples of five peanut m&ms, on average how many do you expect to be red? m&ms with what standard deviation? m&ms

Explanation:

Step1: Identify the binomial - probability formula

The binomial - probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successes, $p$ is the probability of success on a single trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$. Here, $n = 5$ (number of M&M's selected), and $p=0.12$ (probability that a peanut M&M is red).

Step2: Calculate the probability that exactly 4 of 5 are red

First, calculate the combination $C(5,4)=\frac{5!}{4!(5 - 4)!}=\frac{5!}{4!1!}=\frac{5\times4!}{4!×1}=5$. Then, $P(X = 4)=C(5,4)\times(0.12)^{4}\times(1 - 0.12)^{5 - 4}=5\times(0.12)^{4}\times0.88\approx5\times0.00020736\times0.88 = 0.0009$.

Step3: Calculate the probability that exactly 3 of 5 are red

$C(5,3)=\frac{5!}{3!(5 - 3)!}=\frac{5\times4\times3!}{3!×2×1}=10$. $P(X = 3)=C(5,3)\times(0.12)^{3}\times(1 - 0.12)^{5 - 3}=10\times(0.12)^{3}\times(0.88)^{2}=10\times0.001728\times0.7744\approx0.0134$.

Step4: Calculate the probability that 3 or 4 of 5 are red

$P(3\leq X\leq4)=P(X = 3)+P(X = 4)\approx0.0134 + 0.0009=0.0143$.

Step5: Calculate the probability that at most 4 of 5 are red

$P(X\leq4)=1 - P(X = 5)$. $C(5,5)=\frac{5!}{5!(5 - 5)!}=1$, $P(X = 5)=(0.12)^{5}\approx0.0000248832$, so $P(X\leq4)=1-(0.12)^{5}\approx0.9999751168\approx1.0000$.

Step6: Calculate the probability that at least 4 of 5 are red

$P(X\geq4)=P(X = 4)+P(X = 5)\approx0.0009+(0.12)^{5}\approx0.0009 + 0.0000248832=0.0009248832\approx0.0009$.

Step7: Calculate the expected value

The expected value of a binomial - distribution is $E(X)=np$. Here, $n = 5$ and $p = 0.12$, so $E(X)=5\times0.12 = 0.6$.

Step8: Calculate the standard deviation

The standard deviation of a binomial - distribution is $\sigma=\sqrt{np(1 - p)}$. So, $\sigma=\sqrt{5\times0.12\times(1 - 0.12)}=\sqrt{5\times0.12\times0.88}=\sqrt{0.528}\approx0.7266$.

Answer:

Exactly 4 red: 0.0009
3 or 4 red: 0.0143
At most 4 red: 1.0000
At least 4 red: 0.0009
Expected number of red: 0.6
Standard deviation: 0.7266