It has been estimated that about 30% of frozen chicken contain enough salmonella bacteria to cause illness if improperly cooked. A consumer purchases 12 frozen chickens. What is the probability that the consumer will have less than 2 contaminated chickens? Round to 3 decimal places.
The Correct Answer and Explanation is:
This problem can be modeled using a binomial distribution, as there are only two outcomes for each chicken: either it is contaminated or it is not. The binomial distribution formula is:P(X=k)=(nk)pk(1−p)n−kP(X = k) = \binom{n}{k} p^k (1 – p)^{n – k}P(X=k)=(kn)pk(1−p)n−k
Where:
- n=12n = 12n=12 is the number of trials (chickens),
- p=0.30p = 0.30p=0.30 is the probability of a chicken being contaminated,
- kkk is the number of contaminated chickens (which we are interested in, specifically less than 2).
We are asked to find the probability that fewer than 2 chickens are contaminated, i.e., P(X<2)P(X < 2)P(X<2). This is the sum of probabilities for X=0X = 0X=0 and X=1X = 1X=1:P(X<2)=P(X=0)+P(X=1)P(X < 2) = P(X = 0) + P(X = 1)P(X<2)=P(X=0)+P(X=1)
Step 1: Calculate P(X=0)P(X = 0)P(X=0)
The probability of having 0 contaminated chickens is:P(X=0)=(120)(0.30)0(0.70)12=1×1×(0.70)12≈0.0138P(X = 0) = \binom{12}{0} (0.30)^0 (0.70)^{12} = 1 \times 1 \times (0.70)^{12} \approx 0.0138P(X=0)=(012)(0.30)0(0.70)12=1×1×(0.70)12≈0.0138
Step 2: Calculate P(X=1)P(X = 1)P(X=1)
The probability of having exactly 1 contaminated chicken is:P(X=1)=(121)(0.30)1(0.70)11=12×0.30×(0.70)11≈0.0595P(X = 1) = \binom{12}{1} (0.30)^1 (0.70)^{11} = 12 \times 0.30 \times (0.70)^{11} \approx 0.0595P(X=1)=(112)(0.30)1(0.70)11=12×0.30×(0.70)11≈0.0595
Step 3: Add the probabilities
Now, we sum these probabilities to find P(X<2)P(X < 2)P(X<2):P(X<2)=P(X=0)+P(X=1)=0.0138+0.0595=0.0733P(X < 2) = P(X = 0) + P(X = 1) = 0.0138 + 0.0595 = 0.0733P(X<2)=P(X=0)+P(X=1)=0.0138+0.0595=0.0733
Final Answer:
The probability that the consumer will have fewer than 2 contaminated chickens is approximately 0.073.
Explanation:
This solution applies the binomial distribution because the scenario involves multiple independent trials (chickens) where each trial has a fixed probability of success (chicken being contaminated). The key here was identifying that we needed to find the cumulative probability for having fewer than 2 contaminated chickens (i.e., X=0X = 0X=0 and X=1X = 1X=1).
