Question

Of the 98 tiles in a Scrabble set (not including blanks), there are 12 E, 4 S, 4 D and 4 L. what is the probability of drawing the letters of the word SEEDLESS (a) in that order and (b) in any order?

Answer 1

i) When order matters, to form the word seedless, we have to first draw an S from the tiles, probability of selecting S at the first instance = 12/98 ( since there are 12 S's), then probability of selecting an E = 4/97 (since, there are 97 tiles out of which 4 are E's), probability of drawing an E again = 3/96, (since out 96 tiles now, only 3 have E's on them). similarly, we draw the rest of the letters in that order.

Thus, the probability of drawing the letters of the word seedless in that order = 12/98* 4/97*3/96*4/95*4/94*2/93*11/92*10/91

= ( ¹²P₃ * ⁴P₃* ⁴P₁ * ⁴P₁ ) /  ⁹⁸P₈

ii) When order doesn't matter, we can draw 3 S's for 12 S's in ¹²C₃ ways, 3 E's for 4 E's in ⁴C₃ ways, and similarly the D and L can be drawn is ⁴C₁  ways each. The total ways in which we can select 8 tiles without replacement = ⁹⁸C₈

Thus, probability of drawing the letters of the word seedless in any order = = ( ¹²C₃ * ⁴C₃* ⁴C₁ * ⁴C₁ ) /  ⁹⁸C₈