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This is a ratter simple probabilistic problem but i have not seen any similar.

My local lottery works like this:

There are 48 numbers in total (numbered from 1 to 48) You have to pick 5 numbers from 1 to 48 (without repetition) The lottery then draws 5 numbers (without repetition) i) (Easy question): What are the odds of guessing the 5 of the winning numbers? ii) What are the odds of guessing exactly 4 of the 5 winning numbers? iii) What are the odds of guessing exactly 3 of the 5 winning numbers?

After drawing 5 numbers, then there is a 6th number drawn called "silver".

iv) What are the odds of guessing exactly 4 of the 5 winning numbers plus the "silver" number ? (ie: as you have picked 5 numbers, four of them have to be winning numbers and the fifth has to be silver number)

v) What are the odds of guessing exactly 3 of 5 winning number plus the "silver" number ? (ie: as you have picked 5 numbers, three of them have to be winning numbers, one silver number, and you can "miss" one of them.

In all questions, order in which numbers are picked is not important as long as you guess them.

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  • $\begingroup$ Hint (i) : When the first ball is drawn, you have $5$ balls in the lottery out of a possible $48$. Then, after that ball is removed, you have $4$ balls in the lottery out of a possible $47$. Then, after that ball is removed.. $\endgroup$
    – mattos
    Jan 28 '15 at 15:12
  • $\begingroup$ i) is extremly easy, you can guess it intuitively. The problem are the other questions $\endgroup$
    – ctutte
    Jan 28 '15 at 16:13
  • $\begingroup$ Yes, but use (i) as your starting point for (ii), which gives your starting point for (iii) etc. $\endgroup$
    – mattos
    Jan 28 '15 at 16:23
  • $\begingroup$ Well, i have no idea. And i am asking because i do not know tne answers (btw this is not school related) $\endgroup$
    – ctutte
    Jan 28 '15 at 16:33
  • $\begingroup$ It would help if you put what ever work you have done into your question first so I can see where the issues lie. Also, if you could explain to me what have you learnt so far we can take it from there. $\endgroup$
    – mattos
    Jan 28 '15 at 16:53
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There are ${5 \choose n}$ number of ways to choose exactly $n$ balls from the $5$ picked.

There are ${48 - 5 \choose 5 - n}$ number of ways for the other $5 - n$ balls to come out.

There are ${48 \choose 5}$ possible combinations.

Therefore, to pick exactly $4$ balls from $5$ is given by

$$\begin{align} \\ \frac{{5 \choose 4} {43 \choose 1}}{48 \choose 5} &= \frac{5 \cdot 43}{1712304} \\ &= \frac{215}{1712304} \\ &= 0.00012556181 \\ \end{align}$$

To pick exactly $3$ balls from $5$ is given by

$$\begin{align} \\ \frac{{5 \choose 3} {43 \choose 2}}{48 \choose 5} &= \frac{10 \cdot 903}{1712304} \\ &= \frac{9030}{1712304} \\ &= 0.00527359627 \\ \end{align}$$

See how you go with the "silver" questions.

Also, for future reference, lotteries don't pay out according to what the probability of your ticket winning is (none that I know of anyway).

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  • $\begingroup$ Thanks for your response. I knew the odds were very low, but they are extremely low even for the "poor" prizes. $\endgroup$
    – ctutte
    Jan 28 '15 at 17:30
  • $\begingroup$ Can this be the solution to iv): [(5,4)(43,1)/(48,5)]/(42,1) ? $\endgroup$
    – ctutte
    Jan 29 '15 at 14:58
  • $\begingroup$ Try and put it in latex, it's hard to read meta.math.stackexchange.com/questions/5020/… $\endgroup$
    – mattos
    Jan 29 '15 at 14:59
  • $\begingroup$ Can part v) be like this: \begin{align} \\ \frac{{5 \choose 3} {43 \choose 2}} {{48 \choose 5} {5 \choose 2} {43 \choose 1}} \\ \end{align} $\endgroup$
    – ctutte
    Jan 29 '15 at 15:46
  • $\begingroup$ For $(iv)$, it is the probability that you choose exactly $4$ balls, then choose the silver ball ($1$ choice) from your remaining $42$ balls. So your answer was correct. Think about why your part $(v)$ might be wrong.. $\endgroup$
    – mattos
    Jan 30 '15 at 2:41

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