# Thread: Probability Problem: Lottery Tickets

1. ## Probability Problem: Lottery Tickets

Hello everyone!
I have been working on (or rather, looking at) this problem for hours on end, and I am still unable to complete it. For a textbook question, it seems quite difficult to me. What is wrong with me? Any help soon is greatly appreciated. Thank you in advance!

100 tickets are sold in a raffle in which there are 2 prizes. If you buy 5 tickets, what are the chances of you winning:
a.Both prizes?
b.Neither prize?
c.At least one prize?

Hello everyone!
I have been working on (or rather, looking at) this problem for hours on end, and I am still unable to complete it. For a textbook question, it seems quite difficult to me. What is wrong with me? Any help soon is greatly appreciated. Thank you in advance!

100 tickets are sold in a raffle in which there are 2 prizes. If you buy 5 tickets, what are the chances of you winning:
a.Both prizes?
b.Neither prize?
c.At least one prize?

You have 5 chances in 100 to win the first prize.
Then that winning ticket is discarded, so you are left with 4 chances in 99 to win the 2nd prize if you won the first prize.

You have 95 chances in 100 not to win the 1st prize and 94 chances in 99 not to win the 2nd if you didn't win the first.

The probability of winning at least one prize is the probability of winning no prize subtracted from 1.

Maybe you could try calculating those

3. So to find the chances in each case, would I have to multiply or add? That is where I am becoming confused.

100 tickets are sold in a raffle in which there are 2 prizes. If you buy 5 tickets, what are the chances of you winning:

a.Both prizes?
b.Neither prize?
c.At least one prize?

So to find the chances in each case, would I have to multiply or add?

100 tickets are sold in a raffle in which there are 2 prizes. If you buy 5 tickets, what are the chances of you winning:

a.Both prizes?
b.Neither prize?
c.At least one prize?
I cannot tell you the answer to this as it would destroy your comprehension.

You must discover the different ways something can happen.
Calculate the independent probabilities of those ways.
If something can happen in a number of different ways, we add up the probabilities of those seperate ways.

Each of those ways may involve "stages".
For instance, she can win one prize if she wins the first and doesn't win the 2nd,
or she doesn't win the 1st and does win the 2nd.

There are two ways she could win one prize.
Each way contains two stages.

To win 2 prizes she must win the first and the 2nd.
That's the only way that could happen.

For example, suppose we toss a coin.
To get 2 heads would be a 1/4 probability since we can get HH, HT, TH, TT.
Each time we toss it, the probability of a head is 1/2.
If the probability of getting 2 heads is 1/4, how do we combine the two halfs?

5. We multiply! That is what I think. My teacher did give me a hint when I asked him, and he said there may be some multiplication. Or is he giving me a red herring? Hmmm. Is the following working correct?

100 tickets are sold in a raffle in which there are 2 prizes. If you buy 5 tickets, what are the chances of you winning:

a.Both prizes?
5/100 x 4/99
b.Neither prize?
95/100 x 94/100

c.At least one prize?

5/100 ?

We multiply! That is what I think. My teacher did give me a hint when I asked him, and he said there may be some multiplication. Or is he giving me a red herring? Hmmm. Is the following working correct?

100 tickets are sold in a raffle in which there are 2 prizes. If you buy 5 tickets, what are the chances of you winning:

a.Both prizes?
5/100 x 4/99
b.Neither prize?
95/100 x 94/100

c.At least one prize?

5/100 ?
It's far better to know why we multiply or add.

As a guideline, the word "and" in the sentence corresponds to multiplication, since "and" identifies the stages of a way,
the word "or" corresponds to add, since "or" identifies the different ways.

Both prizes... you are correct!

Neither prize.....you are allowing the winner of the first prize to use a "used" ticket, while not allowing it to win!
There should not be 100 tickets in the 2nd draw.

At least 1 means 1 or 2 prizes.
you could sum the probabilities of winning 1 prize and of winning 2 prizes,
or subtract the probability of winning none from 1, since all probabilities sum to 1.

7. Thank you Archie! I swear, the error in (b) was a typographical error; I intended to type 99 rather than 100.

100 tickets are sold in a raffle in which there are 2 prizes. If you buy 5 tickets, what are the chances of you winning:

a. Both prizes?
P (1st prize) x P (2nd prize)
P (1st prize)= 5/100
P (2nd prize)= 4/99
Therefore, 5/100 x 4/99

b. Neither prize?
Therefore, 95/100 x 94/99

c. At least one prize?
5/100 + 4/99

Have I made the correct corrections?

8. Archie, thank you very much for your help. Was my last post correct? For I am going now.

Archie, thank you very much for your help. Was my last post correct? For I am going now.

sorry, I was away and didn't see your last posts til now.

Yes, the first two answers are correct.

For part 3, there are three ways this can happen....

1. win the first prize and not win the second.

2. not win the first prize and win the second.

3. win both prizes.

We sum those three probabilities as they are the three things that can happen.

Each of these "events" of winning at least one prize consists of two stages.

1. the probability of winning the first prize is $\frac{5}{100}$

the probability of not winning the 2nd prize if she won the first prize is $\frac{95}{99}$ because she retains 4 tickets and there are 99 in the draw.

Hence the probability that she will win the second prize only is $\frac{5}{100}\ \frac{95}{99}$

Use the same logic to discover the probability that she does not win the 1st prize and wins the 2nd.

Then calculate the probability that she wins both prizes.

If you build on your work so far, this is do-able.

Finally, note that the probability of her winning "at least" one prize is $1-P(she\ wins\ nothing)$

You've already calculated the probability she wins nothing.

Hence, you should note that you get the same answer whichever way you do part 3.

10. Thank you again, Archie Meade. Despite your absence, I actually did calculate the last answer correctly (eventually).