Question

Distinct four-letter sequences are formed by picking 4 letter tiles from a bag containing 11 different alphabet tiles. Note that the order in which the letters are picked matters. The probability of getting a particular four-letter sequence is .

Answer 1

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.

Answer 2

4 of them
11 choices
slot method
1st slot: 11 choices
2nd slot: 10 choices (1 picked)
3rd slot: 9 choices (1 more picked)
4th slot: 8 choices (1 more picked)

11*10*9*8=7920
7920 ways
so the probabity is
1/7920