Question

Suppose that a bag of scrabble tiles contains 5 Es, 4 As, 3 Ns and 2 Bs. It is my turn and I draw 4 tiles from the bag without replacement. Assume that my draw is uniformly random. Let C be the event that I got two Es, one A and one N. (a) Compute P(C) by imagining that the tiles are drawn one by one as an ordered sample. (b) Compute P(C) by imagining that the tiles are drawn all at once as an unordered sample.

Answer 1

Answer:

Here it is given that bag contains 5E's , 4A's , 3N's & 2B's

So from this total number of balls = 5 + 4 + 3 + 2

= 14

a)

To compute P(C) by imaging that the tiles are drawn one by one as an ordered sample.

We are drawing 2E's . 1A's & 1N's without any replacement

So order is EEAN , EENA , NEEA , AEEN , ANEE , NAEE

So that here we have 6 possible ways

Here when order matters we use the permutations

Now the P(C) value by imagining tiles were drawn as one by one sample

i.e.,

P(C) = (5P2 * 4P1 * 3P4) / 14P4

= 20*4*3 / 24024

= 240/24024

= 0.0099

So P(C) by imagining that the tiles are drawn one by one as an ordered sample = 0.0099

b)

To compute P(C) by imagining that the tiles are drawn all at once as an unordered sample

Here when order doesn't matter we use combinations

So P(C) = 5C2 * 4C1 * 3C1 / 14C4

we know that,

nCr = n! / (n-r)!*r!

P(C) = 10*4*3 / 1001

= 120/1001

= 0.1199

So P(C) by imagining that the tiles are drawn all at once as an unordered sample is 0.1199