Probability Questions

[Diego]

Here there are a some tricky probability Qs

 

I - How many different words with up to 4 letters can be made with the letters A and B?

 

a) 4

b) 16

c) 30

d) 31

e) 32

 

II - What is the probability that you receive a pair when you are dealt two cards from a complete deck?

 

a) 1/17

b) 1/13

c) 1/12

d) 4/17

e) 1/2

 

 

[bazooka76]

For me the first question sounds to be way too general. Given 4 positions and 2 letters A & B. If there is no restriction on the other two letters that form the four letter word, we can say (1)(1)(26)(26) (repition allowed). As I said, this is just one way of looking and interpreting at the question.

 

For the second one.

 

P(getting a pair of cards from a group of 4) = 4/52 * 3/51

P(getting a pair of cards from a pack of cards ie, from 13 sets (1 to A)) = 13*(4/52)*(3/51) = 1/17

(I hope I interpreted the ques correctly though I am not sure)

[roborob]

For the first Q, I would solve it like this. This is the super, duper long way, but I would actually sit there and write it all out.

 

AAAA

AAAB

AABA

ABAA

BAAA

AABB

ABAB

ABBA

BAAB

BBAA

BABA

ABBB

BABB

BBAB

BBBA

BBBB

 

Therefore, totalling 16.

 

Now, for the second Q, I would think of it like this... since they ask for a pair dealt from a deck, that would mean it doesn't matter what you got the first card, the second card would just have to match. So, that would mean that the chances your second card matches your first card would be, 3 out of 51. Does that make any sense at all?

 

Hope this helps,

Rob

 

[Umar]

Yes it makes sense, moreover, you have also got the answer right(1/17).

There must be a fast way to do the first question.

[Dharmin]

I - (a,b)^4 + (a,b)^3 + (a,b)^2 + (a,b) = 16+8+4+2 = 30

II - there are 13 * 4 = 52 cards in complete deck:

so, required probability,

= 13C1 * 4C2 / 52C2

= (13*6) / (26 * 51)

= 1/17

 

Dharmin

[Diego]

Thank you all for your posts,

 

The correct answers are :

 

I - c) = 30

II - a) = 1/17

 

bazooka76, please, explain your approach!? I still don't have clear why you stated this:

"P(getting a pair of cards from a pack of cards ie, from 13 sets (1 to A)) = 13*(4/52)*(3/51) = 1/17"?

 

Rob, your answer makes a lot of sense, thanks.

 

Dharmin, You got them both right, can you explain me how did you arrive to "4C2 / 52C2" on the second one? and by the way thank you for posting also at the gmatclub.

[Dharmin]

Dear Diego, if we need to choose pair of card, we have to select any of the 2 cards from 4 cards. As u know, there are 4 cards of each kind( from A to king)

 

Hope, this will help,

Dharmin

[bazooka76]

Diego,

 

For the second question. In a pack of cards there are 13 kinds of 4 card set such as

1 1 1 1

2 2 2 2

.

.

.

K K K K

A A A A

 

Consider the P(Getting "2 KING(K)s" from a pack of cards) = 4/52 * 3/51

 

But a pair can be either 2 kings OR 2 queens OR 2 4's OR 2 5's etc.. ie total 13 kinds)

= 13 * 4/52 * 3/51 = 1/17

 

Hope this helps.

[Diego]

Now I've got it, thanks a lot Bazooka!!!