- #1

musicgold

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- Homework Statement
- This is not a homework problem. I am trying to understand an example given in a book.

- Relevant Equations
- It is about figuring out the chances of not getting a partial prize.

There is a 3-digit lottery that picks three numbers from the following: 1, 2, 3, 4, 5, 6, 7. The lottery has prizes for the jackpot and deuces. A ticket has a deuce when it has two of the three picked numbers. (see attached pictures if you want more context).

There are 35 possible combinations, and for every jackpot combination, there are 12 possible deuces (12/35). The book talks about a person buying 7 tickets of this lottery and claims that:

Chances of getting no deuces in the seven tickets = 5.3%

Chances of getting all 7 tickets being deuces = 0.1%

I tried to calculate the case of no deuces, but I couldn't get the correct answer. Here is what I did:

Chance of getting the first ticket with non-deuce values and the jackpot = 22/35

Chance of getting the second ticket ... = 21/34

and so on.

I got the following expression

$$ = \frac {22 } {35 } . \frac {21 } {34 } . \frac {20 } {33 } . \frac {19 } {32 } . \frac {18 } {31 } . \frac {17 } {30 } . \frac {16 } {29} $$

$$ = 0.025 $$

What am I missing?

Thanks

There are 35 possible combinations, and for every jackpot combination, there are 12 possible deuces (12/35). The book talks about a person buying 7 tickets of this lottery and claims that:

Chances of getting no deuces in the seven tickets = 5.3%

Chances of getting all 7 tickets being deuces = 0.1%

I tried to calculate the case of no deuces, but I couldn't get the correct answer. Here is what I did:

Chance of getting the first ticket with non-deuce values and the jackpot = 22/35

Chance of getting the second ticket ... = 21/34

and so on.

I got the following expression

$$ = \frac {22 } {35 } . \frac {21 } {34 } . \frac {20 } {33 } . \frac {19 } {32 } . \frac {18 } {31 } . \frac {17 } {30 } . \frac {16 } {29} $$

$$ = 0.025 $$

What am I missing?

Thanks