Answer:
The probability of getting a particular four-letter sequence is \(\frac{1}{7920}\) or 0.0001262
Step-by-step explanation:
We are given that Distinct four-letter sequences are formed by picking 4 letter tiles
Total number of tiles = 11
Now we are also given that the order in which the letters are picked matters.
So, we will use permutation over here .
Formula : \(^nP_r=\frac{n!}{(n-r)!}\)
Now we are given that Distinct four-letter sequences are formed by picking 4 letter tiles from a bag containing 11 different alphabet tiles.
So, n = 11
r = 4
So, The probability of getting a particular four-letter sequence :
= \(\frac{1}{^{11}P_4}\)
= \(\frac{1}{\frac{11!}{(11-4)!}}\)
= \(\frac{1}{\frac{11!}{7!}}\)
= \(\frac{1}{7920}\)
=\(0.0001262\)
Hence the probability of getting a particular four-letter sequence is \(\frac{1}{7920}\) or 0.0001262.
