1) A four-digit number is formed by randomly selecting four digits, without replacement, from the set D = { 1, 2, 3, 4, 5, 6, 7 }. What is the probability that the resulting number is greater than 4200 ?

2) A fair six-sided die is tossed four times, each time noting which numbered face lands up. What is the probability that at least two different numbered faces land up during the four tosses?

I'm not really sure how to approach the first one. And in regards to the second question, I know that I must subtract the complement from 1. Any hints are appreciated.

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    $\begingroup$ For the first one, start thinking in cases: (i) if the first digit is larger than 4, then you don't care about the rest digits. If you fix the first digit, how many choices do you have for the last three digits? (ii) What about if the first digit is 4? Can you consider more cases then? $\endgroup$ – megas Mar 7 '15 at 1:42
  • $\begingroup$ Oh okay so the first digit has to be 4 and the second digit must be 2. Then there are 5 and 4 digits left for the next two slots. I also know that there are 840 total combinations. I'm just not sure how to subtract cases based on their digits. $\endgroup$ – Tyranitator Mar 7 '15 at 1:55

For the first question:

Let's think about just selecting the first digit. There's a probability of $\frac 1 7$ of selecting $7$, $\frac 1 7$ of selecting $6$, and $\frac 1 7$ of selecting $5$. In any of these cases, the resulting number is greater than $4200$. Add up these probabilities: $\frac 1 7 + \frac 1 7 + \frac 1 7 = \frac 3 7$.

Then there's also a $\frac 1 7$ probability of selecting a $4$. Then we need to pick the second digit. If we pick a $2$, then we note that our number will still be greater than $4200$, because picking zeroes for the last two digits is not an option. So the probability of picking one of $\{2,3,5,6,7\}$ is $\frac 5 6$, and any of these will yield a number greater than $4200$. Note that we can't pick $4$ as the second digit because we're selecting from $D$ without replacement.

What's the probability of picking a $4$ and then picking one of $\{2,3,5,6,7\}$? They are independent, so by the multiplication rule we can just multiply them: $\frac 1 7 \cdot \frac 5 6 = \frac 5 {42}$.

So the total probability is $\frac 3 7 + \frac 5 {42} = \frac {18} {42} + \frac {5}{42} = \frac {23}{42}$.

For the second question:

Probability of seeing at least two differently numbered faces $= 1-$ Probability of all four tosses showing the same face.

How do we calculate the probability of all four tosses showing the same face?

Well, suppose we said that all four tosses had to show $6$. The probability of that for a single toss is $\frac 1 6$, and by the rule of multiplication for probabilities of independent events, the chance of tossing four $6$s is $\left(\frac 1 6\right)^4$.

And that's the exact same for all other faces $1,2,3,4,5$. There are six such faces in total, so the probability of all four tosses showing the same face is $6 \cdot \left(\frac 1 6 \right)^4 = 6 \cdot \frac 1 {6^4} = \frac 6 {6^4} = \frac 1 {6^3}$.

(If you want more intuition on this -- draw out the table of probabilities for two rolls of the dice, and note that the diagonal is exactly six events.)

So the probability of seeing at least two numbered faces is $1 - \frac 1 {6^3} = 1 - \frac 1 {216} = \frac {215}{216}$.

  • $\begingroup$ The answer that I've been given for the first question is 23/42. I'm wondering what's going on. You got the second one right though. Thank you. $\endgroup$ – Tyranitator Mar 7 '15 at 2:13
  • $\begingroup$ @Tyranitator You're right, I just tested it by simulation. I made an arithmetic mistake somewhere. Let me check. $\endgroup$ – Newb Mar 7 '15 at 2:23
  • $\begingroup$ @Tyranitator I got it wrong because I thought we were doing this with replacement, rather than without replacement. Fixed. $\endgroup$ – Newb Mar 7 '15 at 2:32
  • $\begingroup$ Oh I see. I really appreciate your help! $\endgroup$ – Tyranitator Mar 7 '15 at 2:35
  • $\begingroup$ Sure, you're welcome! $\endgroup$ – Newb Mar 7 '15 at 2:36

For the first question you can start considering cases. How many $4$-digit numbers are there that are greater than $4200$ whose

  • First digit is equal to $1$? or ($2$ or $3$)
  • First digit is equal to $4$? (Now the second digit matters..)
    • Second digit is equal to $1$?
    • Second digit is equal to $2$?
    • Second digit is equal to $3$? (or $5, 6, 7$) (Why can it not be $4$?)
  • First digit is equal to $5$? (or $6, 7$?)

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