Question

A security alarm requires a four-digit code. The code can use the digits 0–9 and the digits cannot be repeated. What is the approximate probability that the code only contains odd numbers? 0.0005 0.00099 0.012 0.02381

Answer 1

Answer:

0.02381

Step-by-step explanation:

As we know that

We have 10 digits i.e 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0 out of these 5 are odd digits 1, 3, 5, 7, 9.

A security alarm needs a code of four digits that includes only odd values

In the first position of the code is 1 digit from 5 odd digits.

In the second position of the code is 1 digit from 4 digits left.

In the third place it is 1 from 3 digits left

And the last fourth position is 1 from 2 odd digits left

Now we can generate 5: 4: 3: 2 = 120 distinct four-digit code only with odd different numbers in total.

Likewise we can count how much of different codes you can generated from all digits

In the first position of the code it is 1 digit from 10 digits,

In the second position of the code it is 1 digit from 9 digits left,

In the third place it is 8 digits left

And the last fourth position it is 7 digits left.

Now overall you can generate 10:9:8:7 = 5040 that contains different four-digit code by considering different digits.

So, the probability is

$Pr=\dfrac{120}{5040}$

$=\dfrac{1}{42}$

$\approx 0.02381 .$

Answer 2

ANSWER:

D = 0.02381

Step-by-step explanation:

Just took the exam :)

but basically, math. (edg 2020)