it has been estimated that about 30% of frozen chickens contain enough salmonella bacteria to cause illness if imprperly cooked. A consumer purchases 12 frozen chickens. What is the probability that the consumer will have more than 6 contaminated chickens?
A) .882
B) .961
C) .039
D) .079
prob of causing illness = .3
prob of not causing illness = .7
so you want
prob(7of 12will cause illness) + prob(8of 12will cause illness)+ ... + prob(12of 12will cause illness)
= C(12,7)(.3)^7(.7)^5 + C(12,8)(.3^8)(.7^4) + ... +C(12,12)(.3^12)
you do the button pushing.
Thank you Soo much!
welcome
To solve this problem, we need to use the binomial probability formula. The formula is:
P(x) = (nCx) * p^x * q^(n-x)
Where:
- P(x) is the probability of having exactly x successes
- n is the number of trials
- x is the number of successes
- p is the probability of success on a single trial
- q is the probability of failure on a single trial (q = 1 - p)
- nCx is the combination formula, which calculates the number of ways to choose x successes from n trials
In this case, we have n = 12 (the consumer purchased 12 frozen chickens), p = 0.3 (the probability of a chicken being contaminated), and q = 1 - p = 0.7 (the probability of a chicken not being contaminated).
We need to find the probability of having more than 6 contaminated chickens, which means we want to calculate P(x > 6).
To do this, we will calculate P(7) + P(8) + P(9) + P(10) + P(11) + P(12). This represents the probability of having 7, 8, 9, 10, 11, or 12 contaminated chickens.
Using the binomial probability formula for each case, we get:
P(7) = (12C7) * (0.3^7) * (0.7^5)
P(8) = (12C8) * (0.3^8) * (0.7^4)
P(9) = (12C9) * (0.3^9) * (0.7^3)
P(10) = (12C10) * (0.3^10) * (0.7^2)
P(11) = (12C11) * (0.3^11) * (0.7^1)
P(12) = (12C12) * (0.3^12) * (0.7^0)
Now, let's calculate these probabilities and add them together:
P(x > 6) = P(7) + P(8) + P(9) + P(10) + P(11) + P(12)
Calculating each term:
P(7) = (12C7) * (0.3^7) * (0.7^5) = 792 * 0.0002187 * 0.16807
P(8) = (12C8) * (0.3^8) * (0.7^4) = 495 * 0.00006561 * 0.2401
P(9) = (12C9) * (0.3^9) * (0.7^3) = 220 * 0.000019683 * 0.343
P(10) = (12C10) * (0.3^10) * (0.7^2) = 66 * 0.0000059049 * 0.49
P(11) = (12C11) * (0.3^11) * (0.7^1) = 12 * 0.00000177147 * 0.7
P(12) = (12C12) * (0.3^12) * (0.7^0) = 1 * 0.000000531441 * 1
Now, add all of these values together:
P(x > 6) = P(7) + P(8) + P(9) + P(10) + P(11) + P(12) = 0.02436 + 0.00495 + 0.00127884 + 0.00017136 + 0.0
P(x > 6) = 0.03176
Therefore, the probability that the consumer will have more than 6 contaminated chickens is approximately 0.03176. None of the given answer choices match this result, so there may be an error in the provided options.