# A bag contains 8 tiles, each one with a letter from A to H. If 4 tiles are randomly selected from the bag, what is the probablity that the 4 tiles could be arranged to spell either the word "FADE"...

A bag contains 8 tiles, each one with a letter from A to H. If 4 tiles are randomly selected from the bag, what is the probablity that the 4 tiles could be arranged to spell either the word "FADE" or the word "FACE"?

llltkl | College Teacher | (Level 3) Valedictorian

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There are eight lettered tiles in the bag. In order to spell either of the required words, three letters, F, A and E are mandatory and two are optional, D or C.

There are four chances to draw.

The first ‘must pick’ tile (be it F, A or E) can be drawn with a probability of 3/8.

The second ‘must pick’ tile (barring the first, anyone from the remaining 2 required letters) can be drawn with a probability of 2/7.

The third ‘must pick’ tile can be drawn with a probability of 1/6.

The fourth tile (either of D or C, from the remaining letters) can be drawn with a probability of 2/5.

The composite probability is the product of individual probabilities as the events are mutually dependent.

Therefore, overall probability that the four tiles, thus drawn, could be arranged to spell either the word "FADE" or the word "FACE" is 3/8*2/7*1/6*2/5

=1/140

=0.007143

Sources:

aruv | High School Teacher | (Level 2) Valedictorian

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Total number of possible cases =`^8C_4`

The 4 tiles could be arranged to spell either the word "FADE" or the word "FACE".

Let X={F,A,E}

Possible number of case in selection of X from 4 selected letters = 1

Possible number of case in selection of C or D fro 4 selcted

letters= 2

Zaca | Student, Undergraduate | (Level 1) Salutatorian

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We can figure out how many possible combinations (not permutations!) there are, and then find the probability:

(number of combinations that we want) / (number of possible combinations)

How many possible combinations exist, if we choose 4 tiles from 8 tiles?

nCr = `(n!)/((r!)((n-r)!))`

8C4 = `(8!)/((4!)((8-4)!))`

= 70 different combinations

We want two distinct combinations: FACE and FADE

`2/70 = 1/35=2.8%`

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