Answer:
0.1-0.6
Step-by-step explanation:
First let's find the number of the elements in the sample space, that is the total number of codes that can be produced.
The first digit is any of {0, 1, 2...,9}, that is 10 possibilities
the second digit is any of the remaining 9, after having picked one.
and so on...
so in total there are 10*9*8*7 = 5040 codes.
a. What is the probability that the lock code will begin with 5?
Lets fix the first number as 5. Then there are 9 possibilities for the second digit, 8 for the third on and 7 for the last digit.
Thus, there are 1*9*8*7=504 codes which start with 5.
so
P(first digit is five)=
b. What is the probability that the lock code will not contain the number 0?
from the set {0, 1, 2...., } we exclude 0, and we are left with {1, 2, ...9}
from which we can form in total 9*8*7*6 codes which do not contain 0.
P(codes without 0)=n(codes without 0)/n(all codes)=(9*8*7*6)/(10*9*8*7)=6/10=0.6
Answer:
0.1 ; 0.6
[tex]0.1[/tex] is the probability that the lock code will begin with through 9, cannot be repeated, the number 5.
[tex]0.6[/tex] is the probability that the lock code will not contain the number 0.
What is probability?
Probability is the branch of mathematics concerning numerical descriptions of how likely an event is to occur, or how likely it is that a proposition is true. The probability of an event is a number between 0 and 1, where, roughly speaking, 0 indicates impossibility of the event and 1 indicates certainty.
First let's find the number of the elements in the sample space, that is the total number of codes that can be produced.
The first digit is any of {[tex]0, 1, 2...,9[/tex]}, that is 10 possibilities
the second digit is any of the remaining 9, after having picked one.
so in total there are 10×9×8×7 = 5040 codes.
Lets fix the first number as 5.
Then there are 9 possibilities for the second digit, 8 for the third on and 7 for the last digit.
Thus, there are 1×9×8×7=504 codes which start with 5.
The probability that the lock code will begin with through 9, cannot be repeated, the number 5
[tex]=\frac{540}{5400} \\=0.1[/tex]
From the set {[tex]0, 1, 2....,[/tex]} we exclude [tex]0[/tex], and we are left with {[tex]1, 2, ...9[/tex]}
from which we can form in total 9×8×7×6 codes which do not contain 0.
P(codes without 0)=n(codes without 0)/n(all codes)
(9×8×7×6)/(10×9×8×7)
[tex]=\frac{6}{10}[/tex]
[tex]=0.6[/tex]
Hence, the probability that the lock code will not contain the number 0
=[tex]0.6[/tex]
Learn more about probability here:
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