Question

F a number x is in octal form and having zero as unit place then if we change that number x in decimal form then what is the probability that number x having zero as unit place?

Answer 1

If a number x is in octal form and having zero as unit place then if we change that number x in decimal form then what is the probability that number x having zero as unit place?

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1 ANSWER



Saksham Agarwal

Answered Feb 9, 2018

For simplification, initially imagine the octal no. x has 2 digits ( _ _ ). Given, last digit is 0, so our no. becomes ( _ 0 )Now, what can be the values that can fill our blank?
( 0 0), ( 1 0), ( 2 0), ( 3 0), ( 4 0), ( 5 0), ( 6 0), ( 7 0) [remember, all are in hexadecimal]
Which when changed to decimal are respectively:
0, 8, 16, 24, 32, 40, 48, 56
Out of which, how many end with 0 at unit's place? 2
so, for a 2 digit octal no. having 0 in unit's place, 2 out of 8 numbers have 0 in its units place of its respective decimal no.
So, required probability for a 2-digit no. = 2/8 = 1/4

Now, let's imagine the hexadecimal no. is 3 digit. ( _ _ 0 )
The two blanks will have 64 combinations, right?
Out of those 64 combinations, how many combinations will have 0 in its unit's place of its respective decimal representation? (Do some work taking help from the 2 digit part)
(Hint: Write all multiples of 8 ending with 0 until 512 in decimal, then write its respective hexadecimal)
After working, you will get the combinations as:
000, 050, 120, 170, 240, 310, 360, 430, 500, 550, 620, 670, 740 [All nos. are in hexadecimal]
Which are 13 nos. among a combination of 64
So, required probability for a 3 digit no.: 13/64

As the probability depends on the no. of digits in the hexadecimal no., the information in the question If a number x is in octal form and having zero as unit place then if we change that number x in decimal form then what is the probability that number x having zero as unit place?

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1 ANSWER



Saksham Agarwal

Answered Feb 9, 2018

For simplification, initially imagine the octal no. x has 2 digits ( _ _ ). Given, last digit is 0, so our no. becomes ( _ 0 )Now, what can be the values that can fill our blank?
( 0 0), ( 1 0), ( 2 0), ( 3 0), ( 4 0), ( 5 0), ( 6 0), ( 7 0) [remember, all are in hexadecimal]
Which when changed to decimal are respectively:
0, 8, 16, 24, 32, 40, 48, 56
Out of which, how many end with 0 at unit's place? 2
so, for a 2 digit octal no. having 0 in unit's place, 2 out of 8 numbers have 0 in its units place of its respective decimal no.
So, required probability for a 2-digit no. = 2/8 = 1/4

Now, let's imagine the hexadecimal no. is 3 digit. ( _ _ 0 )
The two blanks will have 64 combinations, right?
Out of those 64 combinations, how many combinations will have 0 in its unit's place of its respective decimal representation? (Do some work taking help from the 2 digit part)
(Hint: Write all multiples of 8 ending with 0 until 512 in decimal, then write its respective hexadecimal)
After working, you will get the combinations as:
000, 050, 120, 170, 240, 310, 360, 430, 500, 550, 620, 670, 740 [All nos. are in hexadecimal]
Which are 13 nos. among a combination of 64
So, required probability for a 3 digit no.: 13/64

As the probability depends on the no. of digits in the hexadecimal no., the information in the question is not enough to solve the problem.

not enough to solve the problem.