Question

hi.choose a number X at random from the interval [0,2].What is the probability that
i)the first digit to the right of the decimal is 3?
ii)the second digit is a 7?

Answer 1

i believe it should be \(\frac{1}{10}\) for both

Answer 2

how do you get to that?

Answer 3

the digits can be any one from 1,2,3,4,5....,9,0. So, 1/10.

Answer 4

what are the possibilities for the number X?

Answer 5

is X an integer \(\mathbb Z\)? or any real number \(\mathbb R?\)

Answer 6

i think it is an integer

Answer 7

oh it is a decimal \(\mathbb D \).

the possibilities for \(X\in[0,2]\)

are \(\{0.0,0.1,0.2,0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9,1.0,\cdots\)\[ \quad\qquad \cdots 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0\}\]


there are \(\textbf{two}\) occurrences of each
\[\{ 1, 2, 3, 4, 5, 6, 7, 8, 9, \}\]and \(\textbf {three}\) occurrences of \(\{0,\}\)
as the first digit after the decimal place

this set has ten possible outcomes but not each is equally likely

Answer 8

the probability of a three as the first place after the decimal is
the number of occurrence of three divided by all the possible occurrences
\[P(3)=\frac{2}{21}\]

Answer 9

ok.don't get it entirely bt i'll digest it.thanx @UnkleRhaukus ,@apoorvk ,@satellite73

Answer 10

thanx @apoorvk
@satellite73

Answer 11

Actually it was @UnkleRhaukus really who did a great job here, as he always does!

Answer 12

No worries still ;)

Answer 13

im not entirely sure X is a decimal

Answer 14

i think for ii) ' the second digit ' is again referring to the first to the right digit after the decimal place,

Answer 15

Well, it asks for the 'digit' not what the no. could be. and a digit could only be out of 10 possible options.

WAIT.

If the last digit after the decimal is '0', then it would not be significant right?

Answer 16

hmmm,
that is a good question

Answer 17

thank u all

Answer 18

guess its significance is that u can choose a whole 0 meaning there is no digit after the decimal place.jst like u can choose a whole 1 and 2

Answer 19

maybe i misunderstood the question, but i thought it asked "pick a number between 0 and 2 at random"

Answer 20

not a number with only two decimal places, but any number between 0 and 2
for example \(\frac{1}{e}\) or \(\frac{\pi}{6}\)

Answer 21

in which case the probability that the number will have a 3 to the right of the decimal place means it begins \(0.3...\) or \(1.3...\) and the digits will go on forever. one tenth will begin \(0.0...\) or \(1.0...\) and one tenth will begin \(0.1...\) or \(1.1...\) and so on. therefore if i am interpreting the question correctly, one tenth will begin \(0.3...\) or \(1.3...\)

Answer 22

@UnkleRhaukus i think there might be an error in your calculation as follows. since the number is chosen in the interval \([0,2]\) a continuum, this is a length problem. length of the interval divided by total length. the probability that the number 2 is chosen is zero

Answer 23

i could be wrong

Answer 24

if 2 were excluded wouldn't the interval be \([0,2)\)

Answer 25

the problem is to pick a number at random in the interval [0,2] which is the same as the interval (0,2) as far as probability is concerned. there are an uncountable infinite number of numbers in that interval of length 2, and the probability that any specific number is chosen is zero

Answer 26

for example, if i say "pick a number in the interval [0,2] what is the probability that it is between \(\frac{1}{2}\) and 1, then the solution is the length of the favorable interval, namely \(\frac{1}{4}\) divided by the total length, so the solution is \(\frac{1}{8}\)
the inclusion or exclusion of the endpoints in not important

Answer 27

1/4?

Answer 28

ok that was wrong, i meant \(\frac{1}{2}\) and probability is \(\frac{1}{4}\) a mistake on my part, but idea is clear right?

Answer 29

in other words, on the interval \([0,2]\) \(P(1.3<x<1.4)=P(1.3\leq x\leq 1.4)\)

Answer 30

then why were the end points include in the interval \([0,2]\) instead of \((0,2)\),
or do they mean the same thing in a continuous random distribution?

Answer 31

for a continuous probability it doesn't make any difference.

Answer 32

the probability any specific number is chosen is zero, the solution is the measure of the length of the favorable interval divided by the length of the total interval. the measure of (0, 2) and [0,2] are the same, both intervals have length 2

Answer 33

in fact, since the rational numbers have measure zero, you could not only exclude the end point, but could exclude all rational numbers and the answer would not change

Answer 34

well for my working above i have assumed discrete probability ,
\(X\in \mathbb D\)

Answer 35

i now am becoming convinced that this was a mistake

Answer 36

if the only possible numbers chosen were
\[\{0.0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9\]
\[,1.0,1.1,1.2,1.3,1.4,1.5,1.6,1.7,1.8,1.9,2\}\] then you would be correct for sure

Answer 37

you have 21 discrete numbers, of which 2 have the first digit after the decimal a 3
but i am fairly sure that when it says "pick a number at random" number mean s real number.

Answer 38

ok i concede \( X\in \mathbb R\) \[P(3 \text{ is the first digit to the right of the decomal place})=\frac1{10}\]