Question

Theuniquelettersinoneofourstudent’snamesareASUVandfor

1
another are BEIKLMRY . Say you choose 2 letters from ASUV and 3 letters from

BEIKLMRY. Your goal is not to spell anything; you just want to pick out two letters from the first name and 3 from the second name. Then, what is the probability that you choose exactly 1 vowel from ASUV and 2 consonants from BEIKLMRY?

Answer 1

The unique letters of a student's name are ASUV and another student's name are BEIKLMRY.

The first one contains 2 vowels and 2 consonants.

The second one contains 2 vowels and 6 consonants.

We pick 2 letters from the first one and 3 letters from the second one.

To find the chance that we pick exactly 1 vowel from ASUV and 2 consonants from BEIKLMRY.

Now, from 4 letters in first name, we can select 2 letters in (4 C 2)=6 number of ways.

and, for any of these choices, we can select 3 letters from the 8 letters in the second name in (8 C 3)=56 number of ways.

So, the total number of all possible cases is = 56*6=336.

We can select 1 vowel from 4 letters in first name in (2 C 1)=2 ways, and rest 1 consonant in (2 C 1)=2 ways.

We can select 2 consonants from 8 letters in second name in (6 C 2)=15 ways, and rest 1 vowel in (2 C 1)=2

ways.

So, favourable cases=2*2*15*2=120.

So, the required probability is

=120/336

=5/14.

Thus, the required answer is 5/14.