SOLUTION: Jason has two bags of letters. One of the bags contains each of the six letters in the word ELEVEN, and the other contains the six letters in the word TWELVE, but he cannot remembe

Don't get confused and think it matters whether 

1. He reaches in and draws a first one, then reaches in and draws a second one,
then a third one, and then a fourth one, in sequence.

or whether

2. He just reaches in and grabs any 4 letters in any order, since it's the same
either way.  There is nothing special about the order in which he picks the
four letters, or whether there is no order and he just reaches in one time and
grabs a handful of 4. [It's important to know whether order matters or not.
This is a case where it doesn't.]

The probability in the bag-selecting events A and C are 1/2 each, of course.

The denominator of the probability for the letter selecting events is the
number of ways 4 could be grabbed from either bag. That's 6C4 ways per bag, and
one group of 4 is just as likely as any other.  That makes the denominator
2×(6C4) ways.

Now for the more complicated numerator of the desired probability:
 
To be unable to tell whether he has the "ELEVEN-bag" or the "TWELVE-bag", he
must draw the letters {E,E,L,V} for if he draws an N, a T, a W, or three E's he
will be able to tell, and E,E,L, and V are all that are left when we take away
the letters N, T and W, which he cannot draw.     

So the desired probability is 

P[(A and B) or (C and D)] = P(A)×P(B) + P(C)×P(D)

where events A, B, C, and D are defined as below:

We'll go through these four events one by one to get their probabilities:

A: He selects the ELEVEN bag 

B: He draws {E,E,L,V} from the ELEVEN bag.

C: He selects the TWELVE bag 

D: He draws {E,E,L,V} from the TWELVE bag.

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To find P(A)

P(A) = 1/2, the probability he picks the ELEVEN BAG.  That's easy.

To find P(B)

Easy but not as easy as P(A).

To get the numerator for P(B):
He can pick the 2 E's in 3C2 or 3 ways [Or if you prefer you can look at it as
choosing the 1 E to leave out, which is obviously 3 ways]. He can only pick the
L and V 1 way each. So therefore, 

numerator = 3C2 = 3 
denominator = 2×(6C4)] = 30
P(B) = (3C2)/[2×(6C4)] = 3/30 = 1/10

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P(C) = 1/2, which is the prabability of choosing the TWELVE bag.

---------------------

P(D) is easy because there is only one way to choose the {E,E,L,V} from
the TWELVE-BAG.

numerator = 1
denominator = 2×(6C4) = 30
P(D) = 1/[2×(6C4)] = 1/30

So the desired probability is 

P[(A and B) or (C and D)] =

P(A)×P(B) + P(C)×P(D) = 

(1/2)(1/10) + (1/2)(1/30) =

1/20 + 1/60 =

3/60 + 1/60 =

4/60 =

1/15.

Edwin