A rare mutation only occurs in 1 in every 2048 generations of fruit flies. We can assume that whether or not the mutation occurs is independent of previous generations. (a) Calculate the probability of seeing this mutation at least once in 50 generations. (b) Calculate the probability of seeing this mutation at least once in 3000 generations. (c) Compare the calculation in parts (a) and (b) to the approximation 1 â exp(ânp).

(A) 0.0244

(B) 1 (not 1.47 as is calculated) since probability values are between 0 and 1; 0 and 1 inclusive

Step-by-step explanation:

The rare mutation only occurs in 1 generation, out of every 2048 generations. This implies that the next occurrence will fall in or within the next 2048 generations (2 generations in 4096 generations, will have the rare mutation).

(A) The probability of occurrence of this mutation at least once (at most infinity) in 50 generations of fruit flies will surely be less than, as 50 is less than 2048.

The accurate probability is gotten when 50 is divided by 2048

50÷2048 = 0.0244

(B) The probability of seeing this mutation at least once (at most infinity) in 3000 generations would have been 1.47 but for 3 reasons;

- The full question already tells that the mutation will occur once in every 2048 generations and 3000 is greater than 2048, hence there will be a sure occurrence within 3000 generations.

- Question (b) asks you to calculate the probability of seeing this mutation at least once in 3000 generations so, the probability is 1 (representing full probability).

- In probability theory or statistics, all probability values fall within 0 and 1; with 0 representing no occurrence at all and 1 representing full occurrence.

the correct option is b

step-by-step explanation:

combine all like terms then its just a 2 step equation