?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

2 Answers
Aug 7, 2017

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#

Aug 7, 2017

#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#