Ribonucleic acid (RNA) is a molecule built from sequences of smaller

AAA, CCC , GGG, UUU
Total ways of arranging= 12!/(3!)^4

1) AAA + {CCCGGGUUU}=4C3*9!/(3!)^3
P=4C3*9!/(3!)^3 / 12!/(3!)^4=1/55

2) ACG +{AACCGGUUU}=4C3 3! 9!/((2!)^3.3!)
P=4C3 3! 9!/((2!)^3.3!)/12!/(3!)^4=27/55

There seems to be a typo in "two of other 3 types =
4C3∗3!∗9!/3!2!2!"

Hi Purnank, could you please elaborate & explain what those numbers represent/how you solved this? Your method seems the fastest

See here we have 4 different types.
When all are same that means we have to select one type from 4 so 4C1
and when all are different means we have to select 3 from group of 4 so 4C3
anything else you need?

We are choosing 3 nucleobases from a total of 12, with exactly 3 of each type. The total number of ways to choose any 3 nucleobases from 12 is:
\(12C3\) = \(\frac{ 12!}{10!* 2!}\)

Case 1 - Probability that all 3 nucleobases are the same

• The RNA has 4 types of nucleobases: A, C, G, and U.
• There are exactly 3 of each type in the sequence.

So, the possible choices for picking 3 identical nucleobases are:

• AAA (All Adenine)
• CCC (All Cytosine)
• GGG (All Guanine)
• UUU (All Uracil)

Hence, 4 ways.
So, Probablity that all 3 nucleobases being the same:
= \(\frac{ 4}{12!/(10!* 2!)}\)

= \(\frac{1}{55}\)

Case 2 - Probability that all 3 nucleobases are different
[*]There are 4 types of nucleobases, so we are choosing 3 out of 4.
[*]The number of ways to choose which 3 types we pick from 4 is:

\(4C3\) =4
[*]Once we have chosen which 3 types, we must pick 1 nucleobase from each of those 3 types.
[*]Since each type has 3 choices, the number of ways to do that is:

\(3×3×3=27\)

So, Probability that all 3 nucleobases are different:

= \(\frac{4*27}{12!/(10!* 2!)}\)

= \(\frac{27}{55}\)

karishma could you please share how to approach this?

-> The 12 nucleobases -> (AAA) (GGG) (CCC) (UUU)
-> Collection of all possible sequences consisting of the above 12 nucleobases means all possible arrangements of the above.

How many different arrangements of the 12 nucleobases are possible?

\(\frac{12! }{ 3! 3! 3! 3!}\)

-> One particular sequence (out of the above possible ones) will be selected, and the first 3 nucleobases (for instance, ACG, or AAU) will be detached.

-> Probability that the 3 detached nucleobases are all of the same type

In other words, what is the probability that out of all the possible sequences, the sequence selected was one where the first 3 nucleobases were all of the same type?

(1) Total number of possible sequences = \(\frac{12! }{ 3! 3! 3! 3!}\)

Number of arrangements where the first 3 nucleobases are all of the same type

- First, select the 1 type out of 4 (A,C,G,U) which will form the first 3 nucleobases (e.g. - AAA, or UUU) -> 4C1. There is only one way to arrange these 3 (XXX is one arrangement).
- Now, for the remaining 9 nucleobases, we have a total of \(\frac{9! }{ 3! 3! 3!}\) arrangements.

(2) Number of favorable sequences = \(\frac{4C1 * 9! }{ 3! 3! 3!}\)

P(All the same type) = (2) / (1) = 1/55

-> Probability that the 3 detached nucleobases are of 3 different types

In other words, what is the probability that out of all the possible sequences, the sequence selected was one where the first 3 nucleobases were of 3 different types (Say, AGC, or UAG)?

- First, out of 4 types (A,C,G,U), select the 3 which will be used to form the first 3 nucleobases -> 4C3
- How many arrangements of these 3 are possible? 3!
- Now, for the remaining 9 nucleobases, we have 2 each of the 3 types which were used above and all 3 of the unused type (2,2,2,3).
- Number of arrangements of the 9 nucleobases possible -> \(\frac{9! }{ 2! 2! 2! 3!}\)

(3) Number of favorable sequences = \(\frac{4C3 * 3! * 9! }{ 2! 2! 2! 3!}\)

P(3 different types) = (3) / (1) = 27/55.