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Nice probab. problem!

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by fskilnik@GMATH » Thu Oct 14, 2010 11:44 am
Hi there,

This problem may be done pretty quick and using only GMAT´s level of Maths... insight is necessary, though.

I hope you enjoy it!

Best Regards,
Fábio.

Problem: A fair (6-faces) die is thrown repeatedly, till the number "2" appears for the 7th time. What is the probability that this occurs EXACTLY when you are throwing the die for the 10th time ?

Hint: [spoiler]exactly six "2"´s must appear in the first 9 results...[/spoiler]
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by neerajkumar1_1 » Thu Oct 14, 2010 7:13 pm
is it

1/6 * (1/6)^6 * (5/6)^3 * 9!/(6! 3!)
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by fskilnik@GMATH » Fri Oct 15, 2010 2:51 am
neerajkumar1_1 wrote:is it

1/6 * (1/6)^6 * (5/6)^3 * 9!/(6! 3!)
Yes!! Congrats, neerajkumar1_1.
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by sanju09 » Fri Oct 15, 2010 2:58 am
and...

What is the probability that a letter chosen from the word PROBABILITY is between the letters B and T of the original spelling?


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by fskilnik@GMATH » Fri Oct 15, 2010 11:39 am
sanju09 wrote: What is the probability that a letter chosen from the word PROBABILITY is between the letters B and T of the original spelling?
Nice one, sanju9... not for the weak-of-heart, uh?!

What I like most is the fact that we must take into account the number of repetitions of the letter chosen (from the word PROBABILITY) beforehand! REALLY nice... if nobody comes first, I´ll give my shoot on sunday or monday, ok?!

Have a nice weekend, neerajkumar1_1 too!

Regards,
Fabio.
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by vijchid » Fri Oct 15, 2010 12:59 pm
Is the answer 6/11? Sanju?
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by sanju09 » Fri Oct 15, 2010 9:48 pm
vijchid wrote:Is the answer 6/11? Sanju?
I´ll give my toot on sunday or monday, ok?!
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by bitsho » Sat Oct 16, 2010 1:55 am
The distinct letter in the word "PROBABILITY" - 9 ( P,R,O,B,A,I,L,T,Y) .

so number of ways of choosing a letter from the word = 9c1 = 9

Distinct latters in the word "BABILIT" = 5 (B,A,I,L,T)

Number of ways of choosing a letter from "BABILIT" = 5 .

So required prob= 5/9 . OA pls
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by sanju09 » Sat Oct 16, 2010 2:58 am
bitsho wrote:The distinct letter in the word "PROBABILITY" - 9 ( P,R,O,B,A,I,L,T,Y) .

so number of ways of choosing a letter from the word = 9c1 = 9

Distinct latters in the word "BABILIT" = 5 (B,A,I,L,T)

Number of ways of choosing a letter from "BABILIT" = 5 .

So required prob= 5/9 . OA pls
Please refer to the question again
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by neerajkumar1_1 » Sat Oct 16, 2010 4:02 am
sanju09 wrote:and...

What is the probability that a letter chosen from the word PROBABILITY is between the letters B and T of the original spelling?


~just made
I dont know from where u come up with these questions.. but they are sure fun to solve..

let me see if i can get it..

the word is PROBABILITY

we want the probability that a letter chosen from the original spelling is between B & T
now since there are 2 B's and they are going to create confusion, it will be simpler solving 1 - required..

letters which will not be between B & T will be PROBTY
but since B is already between B & T we can count B... so the not required letters will be PROTY
hence the required prob will be = 1 - 5/11 = 6/11

Let me know sanju09..
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by fskilnik@GMATH » Sun Oct 17, 2010 2:54 pm
sanju09 wrote:What is the probability that a letter chosen from the word PROBABILITY is between the letters B and T of the original spelling?
Funny thing... I understood a COMPLETELY different problem... I will try to create the question stem that I believed was sanju´s idea... and wait for his feedback before giving my (try of) solution...

Let x be one of the letters (randomly) chosen from the word PROBABILITY.
If an anagram of this word is randomly chosen, what is the probability that the chosen letter appears between the 5th and 9th positions (from left to right), both of them included?

Best Regards,
Fabio.
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