Microbes 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 exactly 4
will be mutated?
b. If ten microbes from the colony are sampled, what is the probability that at most 4
will be mutated?
c. If microbe from the colony is examined one by one, what is the probability that the
�rst mutated microbe is the �fth one that is examined?
d. If microbe from the colony is examined one by one, what is the probability that the
�fth mutated microbes is the tenth one that is examined?
To solve these probability questions, we will use the binomial probability formula:
P(X = k) = (nCk) * p^k * (1-p)^(n-k)
Where:
- P(X = k) represents the probability of getting exactly k successes.
- n is the total number of trials.
- k is the number of successful events.
- p is the probability of success for each trial.
- (nCk) denotes the combination of n choose k, which calculates the number of ways to choose k items from a set of n items.
Now let's apply this formula to each question:
a. If ten microbes from the colony are sampled, what is the probability that exactly 4 will be mutated?
To find the probability of exactly 4 mutated microbes out of 10 samples, we use the formula as follows:
P(X = 4) = (10C4) * (1/6)^4 * (5/6)^(10-4)
b. If ten microbes from the colony are sampled, what is the probability that at most 4 will be mutated?
To find the probability of at most 4 mutated microbes out of 10 samples, we need to calculate the probabilities of getting 0, 1, 2, 3, and 4 mutated microbes, and then add them together:
P(X ≤ 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
c. If a microbe from the colony is examined one by one, what is the probability that the first mutated microbe is the fifth one examined?
Since the probability of a microbe being mutated is 1/6, the probability of having the first mutated microbe on the fifth examination is (5/6)^(5-1) * (1/6).
d. If a microbe from the colony is examined one by one, what is the probability that the fifth mutated microbe is the tenth one examined?
The probability of finding the fifth mutated microbe exactly on the tenth examination is (9C4) * (1/6)^5 * (5/6)^(10-5).
Please note that in all these calculations, (nCk) represents the binomial coefficient, which can be calculated as n! / (k! * (n-k)!).