• #1. Exactly 1 zero and rest 2 non-zero ---> {0, x, x}, {x, 0, x}, and {x, x, 0} ---> 3 cases each with 9 possibilities for the 2 non-zero digits ---> 3*9*9 possibilities
  
  #2. Exactly 2 zeroes and rest 1 zero ---> Similar 3 cases each with 9 possibilities for the 1 non-zero digit ---> 3*9 possibilities

varun289 wrote:Thurston wrote an important seven-digit phone number on a napkin, but the last three numbers got smudged. Thurston remembers only that the last three digits contained at least one zero and at least one non-zero integer. If Thurston dials 10 phone numbers by using the readable digits followed by 10 different random combinations of three digits, each with at least one zero and at least one non-zero integer, what is the probability that he will dial the original number correctly?

varun289 wrote:Thurston wrote an important seven-digit phone number on a napkin, but the last three numbers got smudged. Thurston remembers only that the last three digits contained at least one zero and at least one non-zero integer. If Thurston dials 10 phone numbers by using the readable digits followed by 10 different random combinations of three digits, each with at least one zero and at least one non-zero integer, what is the probability that he will dial the original number correctly?

vipulgoyal wrote:experts please tell me where i am wrong

1 zero and 2 non zero
case 1: when 2 non zero are differant
1*9*8 * 3! = 432
+
case 2: when 2 non zero are same
1*9*1 * 3 = 27
+
case 3: 2 zero and 1 non zero
1*1*9 * 3 = 27

required probablity = 10/486

vipulgoyal wrote:my qustion remains same, in case1 all digits are differant 1 zero and two non zero differand digits, thats why i multiplied it by 3!,like the way we arrange ABC among themself

rahulmittal87 wrote:

vipulgoyal wrote:experts please tell me where i am wrong

1 zero and 2 non zero
case 1: when 2 non zero are differant
1*9*8 * 3! = 432
+
case 2: when 2 non zero are same
1*9*1 * 3 = 27
+
case 3: 2 zero and 1 non zero
1*1*9 * 3 = 27

required probablity = 10/486

Vipul, your application is spot-on; However, you have made a slight mistake in case 1 calculation.

case 1: when 2 non zero are differant
1*9*8 * 3! = 432

It should be "3"and not "3!".
So, 3 ( 1 x 9 x 8) = 216 + 27 + 27 = 270
Hence, 10/270 = 1/27

TeonoreFossow wrote:

rahulmittal87 wrote:

vipulgoyal wrote:experts please tell me where i am wrong

1 zero and 2 non zero
case 1: when 2 non zero are differant
1*9*8 * 3! = 432
+
case 2: when 2 non zero are same
1*9*1 * 3 = 27
+
case 3: 2 zero and 1 non zero
1*1*9 * 3 = 27

required probablity = 10/486

Vipul, your application is spot-on; However, you have made a slight mistake in case 1 calculation.

case 1: when 2 non zero are differant
1*9*8 * 3! = 432

It should be "3"and not "3!".
So, 3 ( 1 x 9 x 8) = 216 + 27 + 27 = 270
Hence, 10/270 = 1/27

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