Question

During a recent Scrabble game, I realized that if my opponent has the letter ‘I’ in her tray, she can make a very high value play and I will lose. There are 56 tiles remaining that I can’t see: 7 in my opponent’s tray and 49 in the bag. 4 of them are the letter ‘I.’ What is the probability that my opponent does not have any of the 4 ‘I’ tiles among her 7 tiles? (Hint: What is the probability that my opponent’s first tile is not an ‘I’? Then of the 55 tiles remaining that we haven’t considered yet, what is the probability that her second tile is not an ‘I’? Etc.)

Answer 1

Let X be the number of 'I' tiles in opponent's tray. Here X takes any one of the values 0,1,2,3,4.

There were 56 tiles in the bag 4 of which are 'I' tiles and 7 tiles are on opponent's tray. The drawing of tiles is done without replacement. Each time one draws a tile, either it is a 'I' tile or not.

Thus, XX follows hypergeometric distribution with parameters 56, 4 and 7.

Thus, pmf of X is, (⁴C_x*⁵²C_7-x) /⁵⁶C₇ , x=0(1)4

Therefore, the probability that the opponent does not have any 'I' tile among her 7 tiles is,

P(X=0)

=(⁴C₀*⁵²C₇) /⁵⁶C₇

=0.5769(approximately)

Thus, the required probability is 0.5769.