Question

A bicycle lock has a four-digit code. The possible digits, 0 through 9, cannot be repeated. What is the probability that the lock code will begin with the number 5? . What is the probability that the lock code will not contain the number 0? .

Answer 1

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)=$\frac{n(first -digit- is- 5)}{n(all-codes)}= \frac{1*9*8*7 }{10*9*8*7 }= \frac{1}{10}=0.1$

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

Answer 2

The probability that the lock code will begin with the number 5 is:

• 0.1

The probability that the lock code will not contain the number 0 is:

• 0.6

What is Probability?

This refers to the use of permutations to find the possible occurrence of events.

The total number of elements is 10

Hence, the total codes is 5040.

P(five)= n(first number)/n(all codes)= 1/10= 0.1

Next,

P(codes without zero)= n(codes without zero)/n(all codes)

(9*8*7*6)/(10*9*8*7)=6/10=0.6

=0.6

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