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Assuming I play a Lottery where I can buy a slip with exactly one random number from $1$ to $10$. The Probability of winning this lottery is $\frac{1}{10}$, as only $1$ winning number will be drawn.

Now I add a second lottery game on this Lottery Slip (again $1$ random number from $1$ to $10$). But, when I draw a winning number, this counts for both lotteries.

Example $1$: My Lottery Slip says "Game $1$: $5$ and Game $2$: $8. \implies$ Winning Number is $8\implies$ I won 2nd Lottery.

Example $2$: My Lottery Slip says Game $1$: $8$ and Game $2$: $8. \implies$ Winning Number is $8$ $\implies$ I won both Lotteries.

Example $3$: My Lottery Slip says Game $1$: $1$ and Game $2$: $8. \implies$ Winning Number is $7$ $\implies$ I lost both Lotteries.

What is the probability of at least winning once with a Lottery Slip of $10$ such games? Why?

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  • $\begingroup$ Use mathjax please, and edit, I have done some part. $\endgroup$
    – Sumanta
    Sep 17, 2020 at 12:22
  • $\begingroup$ Sry, I am new to this forum. So I am not very familiar with mathjax. Tried my best now to edit. $\endgroup$
    – Max
    Sep 17, 2020 at 12:31
  • $\begingroup$ Your question seems unclear , how do you define winning a game ? Winning both the draws ? @Max $\endgroup$
    – Crazy over Maths
    Sep 17, 2020 at 12:33
  • $\begingroup$ The thing that confuses me is that, aren't winning numbers tied to the game? I mean, if you have two rounds of games, doesn't that mean that there would be two winning numbers? $\endgroup$
    – Matti P.
    Sep 17, 2020 at 12:37
  • $\begingroup$ Basically, isn't it one lottery and you buy 10 tickets (each with a random number)? $\endgroup$
    – Its_me
    Sep 17, 2020 at 12:38

1 Answer 1

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Your answer of ~65 is almost correct.

However, you have ten of these "double lottery" games. That means the not winning rate is $0.9\cdot 0.9 = 0.81$, since you have two independent picks in one game. This occurs $10$ times, so $0.81^{10}\approx 0.1215$. Subtracting from $1$ gets the answer of $$1-0.1215 = 0.8785 = \boxed{87.9\%}.$$

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  • $\begingroup$ Awsome Thanks :-) $\endgroup$
    – Max
    Sep 17, 2020 at 13:41
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    $\begingroup$ @Max If this helped you, you can click the check mark next to my answer to accept it. Welcome to MSE :) $\endgroup$
    – Sirswagger21
    Sep 17, 2020 at 16:37

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