Random digits appearance

[Dhamnekar Win,odd]

Edited April 3, 2022 by Dhamnekar Win,odd

[John Cuthber]

It's impossible to tell.

What set of digits are the sequence chosen from?

all the digits 0-9

Binary digits 0 and 1  or even hexadecimal digits 0 to F.

[Dhamnekar Win,odd]

1 hour ago, John Cuthber said:

It's impossible to tell.

What set of digits are the sequence chosen from?

all the digits 0-9

Binary digits 0 and 1  or even hexadecimal digits 0 to F.

 All the digits 0-9

[Genady]

I don't think your answer to a) is correct. On one hand, your formula for the case of 5 random digits doesn't make sense to me because it has k in it. On the other hand, the "general terms" answer, (9/10)^k has no justification.

If the a) is indeed incorrect, the rest follows.

[Dhamnekar Win,odd]

59 minutes ago, Genady said:

I don't think your answer to a) is correct. On one hand, your formula for the case of 5 random digits doesn't make sense to me because it has k in it. On the other hand, the "general terms" answer, (9/10)^k has no justification.

If the a) is indeed incorrect, the rest follows.

    Answers to (a) and (b) are correct. [math](0.9)^5 \cdot (0.9)^5 = 0.3486784401 = (0.81)^5 [/math]whereas answer to (c) is  (0.8)⁵  if  k=5 So, there is minor difference of 0.1 in the probability in the answer to (c). I don't understand how did that arise? 

[Genady]

4 minutes ago, Dhamnekar Win,odd said:

    Answers to (a) and (b) are correct. (0.9)5⋅(0.9)5=0.3486784401=(0.81)5 whereas answer to (c) is  (0.8)⁵  if  k=5 So, there is minor difference of 0.1 in the probability in the answer to (c). I don't understand how did that arise? 

I guess I'm missing something. Are we talking about strings of random digits or sets of random digits?

[Dhamnekar Win,odd]

1 hour ago, Genady said:

I guess I'm missing something. Are we talking about strings of random digits or sets of random digits?

Rectified answer to (c) if k =5 :[math]\displaystyle\sum_{k=1}^{5}\binom{5}{k} (02-0.01)^2 (0.81)^{5-k}= 0.6513215599 [/math]= The probability that either 0 or 1 appears.

So, neither 0 nor 1 appears is 1-0.6513215599 = 0.3486784401 = (0.9)⁵ × (0.9)⁵

 

So, we can express event (c) in terms of A and B as A*B or P(A)* P(B).

QED.   

Edited April 3, 2022 by Dhamnekar Win,odd

[Dhamnekar Win,odd]

Rectified answer to (d): 2* (0.9)^k - (0.81)^k