Question

If we create a 5 digit bit string that is randomly generated, all strings equally likely...

1. possibility of string containing three consecutive zeroes?

2. conditional probability of it containing three consecutive zeroes where first number is a one?

Answer 1

Each of the 10 digits can generated for the first position.

Each of the 10 digits can be generated for the second position.

Each of the 10 digits can be generated for the third position.

Each of the 10 digits can be generated for the fourth position.

Each of the 10 digits can be generated for the fifth position.

So number of possible 5 digit bit string that can be randomly generated = 1010101010 = 100000

The string can contain 3 consecutive zeroes in the following ways :-

• If there are 3 consecutive zeroes in the first 3 positions, then the fourth position will not have zero and the fifth position will have any of the 10 digits. So only 0 is generated for the first, second, and third position; any of the 9 digits (excluding 0) can be generated for the fourth position; and any of the 10 digits can be generated for the fifth position. Number of such 5 digit bit string that can be randomly generated = 111910 = 90
• If there are 3 consecutive zeroes in the middle 3 positions, then the first and the fifth position will not have zero.. So only 0 is generated for the second, third and fourth position; any of the 9 digits (excluding 0) can be generated for the first and the fifth position. Number of such 5 digit bit string that can be randomly generated = 91119 = 81
• If there are 3 consecutive zeroes in the last 3 positions, then the second position will not have zero and the first position will have any of the 10 digits. So only 0 is generated for the third, fourth, and fifth position; any of the 9 digits (excluding 0) can be generated for the second position; and any of the 10 digits can be generated for the first position. Number of such 5 digit bit string that can be randomly generated = 109111 = 90

So possibility of string containing three consecutive zeroes

= (90+81+90)/100000 = 0.00261

The string can contain 3 consecutive zeroes and 1 as the first number in the following ways :-

• If 1 is the first number and there are 3 consecutive zeroes in the middle 3 positions, then fifth position will not have zero.. So only 1 is generated for the first position and only 0 is generated for the second, third and fourth position; any of the 9 digits (excluding 0) can be generated for the fifth position. Number of such 5 digit bit string that can be randomly generated = 11119 = 9
• If 1 is the first number and there are 3 consecutive zeroes in the last 3 positions, then the second position will not have zero. So only 1 is generated for the first position and only 0 is generated for the third, fourth, and fifth position; any of the 9 digits (excluding 0) can be generated for the second position. Number of such 5 digit bit string that can be randomly generated = 19111 = 9

So possibility of string containing three consecutive zeroes and 1 as the first number

= (9+9)/100000 = 0.00018

If 1 is the first number; then each of the second, third, fourth and fifth position can contain any of the 10 digits. So only 1 is generated for the first position and any of the 10 digits can be generated for each of the second, third, fourth and fifth position. Number of such 5 digit bit string that can be randomly generated = 110101010 = 10000

So possibility of string where the first number is one

= 10000/100000 = 0.1

So conditional probability of it containing three consecutive zeroes where first number is a one

= 0.00018/0.1 = 0.0018