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29 Oct 2021, 08:15
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
68% (01:46) correct
32%
(02:32)
wrong
based on 136
sessions
History
Date
Time
Result
Not Attempted Yet
What is the probability that all digits of a certain 5-digit password are prime numbers or zero?
A. \(\frac{ 1}{32}\)
B \(\frac{1 }{16}\)
C. \(\frac{1}{ 8 }\)
D. \(\frac{1}{ 5 }\)
E. \(\frac{2}{ 5 }\)
A. \(\frac{ 1}{32}\)
B \(\frac{1 }{16}\)
C. \(\frac{1}{ 8 }\)
D. \(\frac{1}{ 5 }\)
E. \(\frac{2}{ 5 }\)
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29 Oct 2021, 12:23
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Option (A)
Number available 0 to 9 .
Total = 10 numbers .
The code is of 5 numbers .
first number can be selected in 10c1 ways .
Since the numbers are not unique and repetition is allowed we can select the second number in 10c1 ways and so on .
So total possible ways in which this code can be made is (10c1)(10c1)(10c1)(10c1)(10c1).
Now the possibility of finding the code with only prime numbers and zero .
So total numbers to select from 0,2,3,5,7 .
Total numbers 5 .
The first number can be selected in 5c1 ways again repetition allowed so second number can be selected in 5c1 ways and so on .
So numbers of ways in which the code with odd numbers is selected is (5c1)(5c1)(5c1)(5c1)(5c1) .
Probability = (number of required outcomes )/ (total possible outcomes)
= (5c1)(5c1)(5c1)(5c1)(5c1) / (10c1)(10c1)(10c1)(10c1)(10c1)
= 1/32
Number available 0 to 9 .
Total = 10 numbers .
The code is of 5 numbers .
first number can be selected in 10c1 ways .
Since the numbers are not unique and repetition is allowed we can select the second number in 10c1 ways and so on .
So total possible ways in which this code can be made is (10c1)(10c1)(10c1)(10c1)(10c1).
Now the possibility of finding the code with only prime numbers and zero .
So total numbers to select from 0,2,3,5,7 .
Total numbers 5 .
The first number can be selected in 5c1 ways again repetition allowed so second number can be selected in 5c1 ways and so on .
So numbers of ways in which the code with odd numbers is selected is (5c1)(5c1)(5c1)(5c1)(5c1) .
Probability = (number of required outcomes )/ (total possible outcomes)
= (5c1)(5c1)(5c1)(5c1)(5c1) / (10c1)(10c1)(10c1)(10c1)(10c1)
= 1/32
29 Oct 2021, 18:58
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Bunuel
There are 4 prime numbers 2,3,5 and 7. If the digits can be 0 or prime, we have 5 digits to choose from. 1st place can be filled in 5 ways,2nd in 5 ways and so on.
Total ways of filling 5 digits - 5^5
Total ways of selecting 5 digits without any restrictions will be 10^5
Probability = 5^5/10^5 = 1/32
Option A
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