icrobes in a certain colony exposed to radiation have a 1/6 chance of being
mutated.
a. If ten microbes from the colony are sampled, what is the probability that at most 4 will be mutated?
b. If microbe from the colony is examined one by one, what is the probability that the first mutated microbe is the �fifth one that is examined?
c. If microbe from the colony is examined one by one, what is the probability that the �fifth mutated microbes is the tenth one that is examined?
To solve these probability problems, we will use the concept of binomial probability. Binomial probability is used when there are two possible outcomes (mutated or not mutated) in a fixed number of trials (examining or sampling microbes). The probability of success in each trial (mutation occurring) remains constant.
a. Probability that at most 4 microbes will be mutated:
To find the probability that at most 4 microbes will be mutated out of 10 sampled, we need to calculate the individual probabilities for 0, 1, 2, 3, and 4 mutated microbes, and then add them together.
The formula for calculating the probability of x successes in n trials is:
P(x) = (nCx) * p^x * q^(n-x)
where:
P(x) is the probability of x successes
n is the total number of trials (10 microbes sampled)
x is the number of successes (0, 1, 2, 3, or 4 microbes mutated)
p is the probability of success in each trial (1/6 chance of mutation)
q is the probability of failure in each trial (1 - p)
Using this formula, we can calculate the probabilities for 0 to 4 mutated microbes:
P(0) = (10C0) * (1/6)^0 * (5/6)^10
P(1) = (10C1) * (1/6)^1 * (5/6)^9
P(2) = (10C2) * (1/6)^2 * (5/6)^8
P(3) = (10C3) * (1/6)^3 * (5/6)^7
P(4) = (10C4) * (1/6)^4 * (5/6)^6
Finally, we sum up these probabilities to get the probability that at most 4 microbes will be mutated:
P(at most 4) = P(0) + P(1) + P(2) + P(3) + P(4)
b. Probability that the first mutated microbe is the fifth one examined:
To calculate this probability, we need to consider that the first four microbes examined are not mutated, and the fifth one is mutated.
The probability that a microbe is mutated is 1/6, and the probability that it is not mutated is 5/6. Therefore, the probability that a specific order of microbes occurs (4 not mutated, 1 mutated) is:
(5/6)^4 * (1/6)
c. Probability that the fifth mutated microbe is the tenth one examined:
In this case, we consider that the first nine microbes are not mutated, and the tenth one is mutated.
Using the same probability values as before, the probability of this specific order is:
(5/6)^9 * (1/6)
Remember, these probabilities are applicable if the conditions and assumptions hold true for the problem scenario.