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31 May 2024, 03:07
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All the same type: \(\frac{1}{55}\)
3 different types: \(\frac{27}{55}\)
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Ribonucleic acid (RNA) is a molecule built from sequences of smaller molecules called nucleobases. RNA nucleobases are of 4 different types: adenine (A), cytosine (C), guanine (G), and uracil (U). Consider the collection of all possible RNA sequences consisting of 12 nucleobases, 3 of each type. An RNA sequence will be selected at random from this collection, and the first 3 nucleobases of the sequence will be detached from the sequence.
In the table, select the probability that the 3 nucleobases are all of the same type, and select the probability that they are of 3 different types. Make only two selections, one in each column.
In the table, select the probability that the 3 nucleobases are all of the same type, and select the probability that they are of 3 different types. Make only two selections, one in each column.
ID: 101100
| All the same type | 3 different types | |
| \(\frac{1}{55}\) | ||
| \(\frac{9}{220}\) | ||
| \(\frac{9}{110}\) | ||
| \(\frac{9}{55}\) | ||
| \(\frac{27}{55}\) |
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Official Answer
All the same type: \(\frac{1}{55}\)
3 different types: \(\frac{27}{55}\)
31 May 2024, 05:10
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guddo
Each RNA sequence will be made from A,A,A,C,C,C,G,G,G,U,U, and U.
Total ways of arranging, say n things, when x are similar, y are similar and of another type: n!/x!y!
We divide by x! and y! as these similar items can be arranged within themselves in x! and y! ways, and taking them separately gives repetitions.
Total ways = \(\frac{12!}{3!3!3!3!} \)
(I) First 3 nucleobases are all of the same type: So, first three places can be taken by any of the four, so 4 ways, while remaining 9 places can be filled by other 3 types = \(\frac{4*9!}{3!3!3!}\)
P =\( \frac{\frac{4*9!}{3!3!3!}}{\frac{12!}{3!3!3!3! }}= \frac{4*3!}{12*11*10 }= \frac{1}{55}\)
(II) First 3 nucleobases are of three different types: So, first three places can be taken by any three of the four, so 4C3 ways or 4C3*3! when arranged, while remaining 9 places can be filled by three of one kind and two of other 3 types = \(\frac{4C3*3!*9!}{3!2!2!}\)
P =\( \frac{\frac{4!*9!}{3!2!2!2!}}{\frac{12!}{3!3!3!3! }}= \frac{4!*3*3*3}{12*11*10 }= \frac{27}{55}\)
_______________________________________________
Just adding a faster method ...
There are three of each type, so in total 12 nucleobases say A1, A2, A3, B1, B2......U3.
1) All three are of same type..
Choose any of 3 As from A1, A2, and A3...So 3C3 or 1 way..
Similarly 1 way for each of Cs, Gs and Us.
Thus total 4 ways.. and each such selection can be arranged in 3! ways
Total ways = 12*11*10
P = 4*3*2/12*11*10 or 1/55
2) All three are different
Choose any of the 12 (say A) for first place.
Next leaving out that (A) here, choose from remaining 9, say C here
Finally choose from 6 of Gs and Us left..
Total 12*9*6
Total ways 12*11*10
P = 12*9*6/12*11*10 = 27/55
06 Aug 2024, 08:40
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guddo
all of the same type = \(\frac{4C1*3C3}{12C3}=\frac{1}{55}\)
3 different types = \(\frac{4C3*3C1*3C1*3C1}{12C3} = \frac{27}{55}\)
