Thread: Whats the probability the parameters of my random sample match those of the whole?

1. Whats the probability the parameters of my random sample match those of the whole?

I have been posed a problem, which i am struggling with:

If I have a box of 2,808 lottery style tickets with winning ticket /denominations as follows:

Tickets in Box 2808

Number of Winning Tickets Prize
2 £300
2 £100
2 £50
20 £15
80 £5
222 £2

If I were to take a bundle of 59 tickets completely at random from the box of 2,808.

How many winning £2 prize tickets in all probability could I expect to have?
Now you could say well ALL 59 tickets could be £2 winning tickets, if you were really lucky, as there are 222 x £2 winning tickets in the box....but I wonder if that's fair to assume

Have I oversimplified the problem by assuming 222 (£2 tickets) / 2808 (tickets in the box) = 10.6% chance

Therefore, 10.6% x 59 (tickets in my randomly bundle) = I could have expected there to be 6 x £2 tickets?

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Of the 2,808 tickets in the box, only 328 tickets are 'winning' tickets (2+2+2+20+80+222) or 15.7% of the total,

the remaining 1,760 tickets (2808 - 328) or 84.3% of the total are non-winners

Or does my calculation need more work e.g as follows, calculating each tickets probability, and the total tickets reducing each time...

for example looking at the £300 prize, there are 2 winning tickets in a box of 2,808, so taking 1 ticket randomly = odds of 2 tickets out of 2808 total tickets (or 0.07%) chance of winning,
and then assuming that ticket is then gone, the odds change as the next chance of taking a £300 prize ticket = 1/2807 (or 0.04%) etc..

2. Re: Whats the probability the parameters of my random sample match those of the whole

Let's assume one has a winning ticket. The probability that the winning ticket is £2 winning ticket is 222/328. The number of winning tickets you can get out off the 59 follows a hypergeometric distribution. This is because the population of tickets can be divided into winning (324) and non-winning ticket (2808-324 = 2484) subgroups. Prob(getting x number of winning tickets from 59 tickets) = 324Cx.2484C(59-x) where x is a random variable, C is combinatorial sign, and the . sign represents multiplication. E(x)= expected number of winning tickets which is given by the formula 59.324/2808. Now expected number of winning tickets multiplied by the probability that the winning ticket is £2 winning ticket is 59.324/2808.222/328. Simplify this expression to get 5 approx. oparairoegbu.

3. Re: Whats the probability the parameters of my random sample match those of the whole

In my earlier post I wrote "Prob(getting x number of winning tickets from 59 tickets) = 324Cx.2484C(59-x) where x is a random variable, C is combinatorial sign, and the . sign represents multiplication. Sorry, the hypergeometric distribution formula was incomplete. The correct formula is 324Cx.2484C(59-x)/2808C59.
oparairoegbu

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