Question

Blank/Wild: 2 tiles. 0 pointsA: 9 tiles,1 pointB: 2 tiles,3 pointsC: 2 tiles,3 pointsD: 4 tiles,2 pointsE: 12 tiles,1 pointF: 2 tiles,4 pointsG: 3 tiles,2 pointsH: 2 tiles,4 pointsI: 9 tiles,1 pointJ: 1 tile,8 pointsK: 1 tile,5 pointsL: 4 tiles,1 pointM: 2 tiles,3 pointsN: 6 tiles,1 pointO: 8 tiles,1 point P: 2 tiles,3 pointsQ: 1 tile,10 pointsR: 6 tiles,1 pointS: 4 tiles,1 pointT: 6 tiles,1 pointU: 4 tiles,1 pointV: 2 tiles,4 pointsW: 2 tiles.4 pointsX: 1 tile,8 pointsY: 2 tiles,4 pointsZ: 1 tile,10 points4. Above is a graphic showing the distribution of tiles in Scrabble, showing how many of each letter there are as well as how many points there are worth. Note that there are 100 tiles total.(b) What is the probability of picking two vowels in a row, if you do not replace the first oneyou picked? (Do not count Y as a vowel.)

Answer 1

`0.174`

Step-by-step explanation:

Here, we want to get the probability of picking two vowels in a row

The letters of a vowel are:

A , E , I , O , U

Let us sum up the number of tiles for all vowels:

We have that as the sum of the number of tiles of each of the vowels

Mathematically, we have that as:

9 + 12 + 9 + 8 + 4 = 42

The probability of picking a vowel is the number of vowel tiles divided by the total number of tiles

We have that as:

`(42)/(100)`

Now, we are not replacing the first vowel and we are going for another pick

The total number of vowel tiles is now 41 while the total number of tiles is 99

The probability of picking a vowel now will be:

`(41)/(99)`

Finally, we have the probability of picking two vowels in a row as the product of the two probabilities above

Mathematically, we have that as:

`(41)/(99)*(42)/(100)\text{ = 0.174}`