Dhamnekar Win,odd Posted April 3, 2022 Share Posted April 3, 2022 (edited) Edited April 3, 2022 by Dhamnekar Win,odd Link to comment Share on other sites More sharing options...
John Cuthber Posted April 3, 2022 Share Posted April 3, 2022 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. Link to comment Share on other sites More sharing options...
Dhamnekar Win,odd Posted April 3, 2022 Author Share Posted April 3, 2022 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 Link to comment Share on other sites More sharing options...
Genady Posted April 3, 2022 Share Posted April 3, 2022 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. Link to comment Share on other sites More sharing options...
Dhamnekar Win,odd Posted April 3, 2022 Author Share Posted April 3, 2022 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? Link to comment Share on other sites More sharing options...
Genady Posted April 3, 2022 Share Posted April 3, 2022 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? Link to comment Share on other sites More sharing options...
Dhamnekar Win,odd Posted April 3, 2022 Author Share Posted April 3, 2022 (edited) 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 Link to comment Share on other sites More sharing options...
Dhamnekar Win,odd Posted April 4, 2022 Author Share Posted April 4, 2022 Rectified answer to (d): 2* (0.9)k - (0.81)k Link to comment Share on other sites More sharing options...
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