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Probability doubt

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by sachin_yadav » Wed Dec 08, 2010 7:36 pm
Hi All,

Came across a question from "Veritas prep".

A telephone number contains 10 digits, including a 3-digit area code. Bob remembers the area code and the next 5 digits of the number. He also remembers that the remaining digits are not 0, 1, 2, 5, or 7. If bob tries to find the number by guessing the remaining digits at random, the probability that he will be able to find the correct number in at most 2 attempts is closest to which of the following ?

(A) 1/625
(B) 2/625
(C) 4/625
(D) 25/625
(E) 50/625


Thanks
Sachin
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by shovan85 » Wed Dec 08, 2010 8:10 pm
Numbers left to fill last two digits are 3,4,6,8,9.
So, Total 5 numbers are left to be chosen.

Probability of choosing right numbers in two places
= Probability of choosing right number in first place * Probability of choosing right number in second place
=1/5 * 1/5
= 1/25

Thus, Probability of not choosing right numbers in two places = 1-1/25 = 24/25

Two attempts and it has to succeed thus

1) 1st Attempt: Wrong, 2nd Attempt: Correct
So, P = 24/25 * 1/25 = 24/625

2) 1st Attempt: Correct
So, P = 1/25

Thus requiredd probability 24/625 + 1/25 = 49/625

IMO E Nearest One
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by beat_gmat_09 » Wed Dec 08, 2010 8:24 pm
Numbers to fill - 3,4,6,8,9
5 can be filled in 5*5 = 25 ways, assuming repetition.

Probability of getting in right in 2nd attempt is just as selecting 1 option from 25 Proba = 1/25
Probab of getting right in 1st attempt = 1/25

Total probab = 1/25 + 1/25 = 2/25
2/25 *25/25 = 50/625
IMO E
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by Night reader » Wed Dec 08, 2010 8:45 pm
sachin_yadav wrote:Hi All,

Came across a question from "Veritas prep".

A telephone number contains 10 digits, including a 3-digit area code. Bob remembers the area code and the next 5 digits of the number. He also remembers that the remaining digits are not 0, 1, 2, 5, or 7. If bob tries to find the number by guessing the remaining digits at random, the probability that he will be able to find the correct number in at most 2 attempts is closest to which of the following ?

(A) 1/625
(B) 2/625
(C) 4/625
(D) 25/625
(E) 50/625


Thanks
Sachin
5 digits are left =>
there are 5 ways to arrange 5 digits for each of two units (last) in the number

___ ___ => 5*5 => 25
P(1st attempt) = 1/25

P(1st attempt wrong) = 1-1/25 = 24/25

P(2nd attempt) = 1/24

max P(correct @2nd attempt) = (24/25)*(1/24) =24/600

Combining two probabilities 24/600 + 1/25 = (24+24)/600 = 48/600 = 12.5

50/625 = 48/600

Answer E is correct
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by sachin_yadav » Wed Dec 08, 2010 10:36 pm
Thank you guys. I really appreciate for the help, but probability is definitely a nightmare for me.

Regards,
Sachin
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