Question

?Ten slips of paper are each marked with a different number (from 0 to 9) and placed in a basket. A slip is pulled out, its number is recored (in the order it is drawn), and the slip is replaced. This is done 4 times.

How to find the probabilities that the follow (numbers) are formed?

a) 1234
b) a number which starts with 5
c) a number with no repetition of digits.
d) a number that does not contain 2,3, or 4

Answer 1

As explained below.

Explanation:

There are `10` different digits. The probability of drawing any number is `1/10`

a) To form the number `1234`, no choice is allowed.
The numbers must be drawn in this order.

`P(1) xx P(2) xx P(3) xx P(4) = 1/10xx 1/10xx1/10xx1/10`

`= 1/(10,000)`

The total number of ways of drawing `4` cards is:

`10xx10xx10xx10 = 10,000`

b) To form a number which start with `5`, there is only `1` digit allowed for the first card.

However, the numbers which follow can be any of the `10` digits.

`P(5N N N) = 1000/10000 =1/10`
This can also be calculated as `1/10 xx10/10xx10/10xx10/10 =1/10`

c) If no repetition is allowed, then the number of choices drops with each successive draw.
There are `10` choices for the first, then, `9`, then `8`, then `7`.

Number of ways `= 10xx9xx8xx7 = 5040`
(This allows for the first digit to be `0`)

`10/10 xx9/10xx8/10xx7/10 = 5040/10000`

If the first digit may not be `0`,
then the number of ways `= 9xx9xx8xx7 =4536`

`9/10xx9/10xx8/10xx7/10 = 4536/10000`

`P("no digit repeating") = 5040/10000 or 4536/10000`

d) If a number may not contain `2,3,4` then there are only `7` choices.

Number of ways is `7xx7xx7xx7 = 2401`

If the first digit may be `0`

`7/10xx7/10xx7/10xx7/10 = 2401/10000`

`P("not 2,3,4") = 2401/10000`

If the first digit may not be `0`, then
the number of ways `6xx7xx7xx7 = 2058`

`P("not 2,3,4") = 2058/10000`

Answer 2

`a. 1/10000=0.0001`

`b. 1/10=0.1`

`c. 504/1000=63/125=0.504`

`d. 2401/10000=0.2401`

Explanation:

a. `P("Any card")=1/10`

Therefore, `P(1)=1/10`

`P(12)=(1/10)*(1/10)=(1/10)^2=1/100`

`P(123)=(1/10)*(1/10)*(1/10)=(1/10)^3=1/1000`

`P(1234)=(1/10)*(1/10)*(1/10)*(1/10)=(1/10)^4=1/10000`

b. `P(5)=1/10`

`SigmaP(5"x")=(1/10)*(10/10)=1/10*1=1/10`

`SigmaP(5"xx")=(1/10)*(10/10)*(10/10)=1/10*1*1=1/10`

`SigmaP(5"xxx")=(1/10)*(10/10)*(10/10)*(10/10)=1/10*1*1*1=1/10`

c. With one card: `P("No repeatition")=10/10=1`

With two cards: `P("No repeatition")=10/10*9/10=0.9`

Explanation: with one card, it is impossible to repeat numbers, so `P("No repeatition")=1`. However, with two cards, there are 100 combinations, however, for every card, only 9 involve no repeatition. So, `P("No repeatition")=(9*10)/(10*10)=90/100=0.9`

With three cards: `P("No repeatition")=10/10*9/10*8/10=0.72`

With four cards: `P("No repeatition")=10/10*9/10*8/10*7/19=0.504`

d. `P("Not " 2, 3 or 4)=7/10`

`P(4("Not " 2, 3 or 4))=(7/10)^4=2401/10000`