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Here there are a some tricky probability Qs I - How many

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Intern
Joined: 20 May 2004
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Here there are a some tricky probability Qs I - How many [#permalink]  22 May 2004, 08:32
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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
Senior Manager
Joined: 02 Mar 2004
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1. |{A, B}^4| + |{A, B}^3|+|{A, B}^2|+|{A, B}| = 16+8+4+2 =30
2. I dont understand the question--deck etc.
Senior Manager
Joined: 06 Dec 2003
Posts: 366
Location: India
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Re: PS Probability Qs !!! [#permalink]  22 May 2004, 23:23
Diego wrote:
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

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
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hallelujah1234, you got the first one right, I am new to the GMAT, can you explain me your approach for your answer?

SVP
Joined: 30 Oct 2003
Posts: 1794
Location: NewJersey USA
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if you have digits 0 to 9 (10digits ) and there can be 100 different 2 digit numbers. This is same as 10 * 10
If you have 3 digits number then you have 10 * 10 * 10 variations

In your problem if you want a word with one letter then you have 2 variations A and B
words with 2 letters = 2 * 2
words with 3 letters = 2 * 2 * 2
worda with 4 letters = 2 * 2 * 2 * 2
Add all these you will have 16+8+4+2 = 30
Senior Manager
Joined: 07 Oct 2003
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Re: PS Probability Qs !!! [#permalink]  12 Jun 2004, 14:40
Dharmin wrote:
Diego wrote:
Here there are a some tricky probability Qs

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

II - there are 13 * 4 = 52 cards in complete deck:
so, required probability,
= 13C1 * 4C2 / 52C2
= (13*6) / (26 * 51)
= 1/17

Dharmin

Here's what I don't get -- there are 26 ways to get a pair in a deck of cards. Why are we not computing 26C1* 4/52 * 3/51?

?
Manager
Joined: 02 Jun 2004
Posts: 154
Location: san jose,ca
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lastochka,

there are not 26 ways,actually 13*6=78 ways.
Let me see if I can explain this in simple words.

For each number you have four cards(one hearts,one clubs,one dice and once spade).You can pick them in 6 ways ({hearts,clubs},{hearts,dice},{hearts,spade},{clubs,dice},{clubs,spade},{dice,spade}). You donot need to calculate like this all the time. Picking 2 items from 4 items can be done in 4C2=6 ways.

All this is for one number.We have 13 numbers like this (A,2, 3, ,...J,Q,K).So you can pick a pair in 13*6 ways.

Hope this helps.
_________________

GS
No excuses - Need 750!!!

Senior Manager
Joined: 07 Oct 2003
Posts: 358
Location: Manhattan
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great explanation goalstanford. thanks.
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