Determining which Probability Distrubitive Function to use

Question

I have two questions that I am to solve as practice. I am having difficulties with determining which probability function to use.

The firstquestion is:

Todd decides to keep buying a lottery ticket each week until he has 4 winners (of some prize). Suppose 30% of the tickets win some prize. Find the probability he will have to buy 10 tickets.

For this question, I should apply the Binomial distribution function. let X be the number of lottery tickets bought

X ~ $Binomial(10, 0.3)$

$P(x) = ((10) nCr (x))(0.3)^x(0.7)^(10-x)$

$P(4) = (10 nCr 4) * (0.3)^4*(0.7)^6=0.200$

However, the solution is 0.08.

The second question is:

An oil company runs a contest in which there are 500,000 tickets; a motorist receives one ticket with each fill-up of gasoline, and 500 of the tickets are winners. If a motorist has ten fill-ups during the contest, what is the probability that he or she wins at least one prize?

Here my random variable would the number of tickets received. I would use the Hypergeometeric distributive function because we have a finite set.

I attempted to solve the above question, but the solution is not what I am getting.

Here is my attempted solution:

let X be the number of tickets received through gas fill-ups

X ~ $Hypergometeric(500 000, 500, 10)$

$P(x) = (500 nCr x)(450000 nCr 10-x)/(500000 nCr 10)$, Where x is an element of [0,10]

Since the question explicitly asks for at least 1 ticket, so:

$P(X >= 1) = 1 - P(0) = 1 - [(500 nCr 0)(450000 nCr 10)/(500000 nCr 10)]= 0.651$

However, the answer is 0.010. I know I am doing something terribly wrong, but I am not too sure what it is.

I am also sorry for not properly using Latex.

Answer 1

For the first question, we want exactly $3$ successes in the first $9$ trials, and success on the $10$-th. The probability is $$\binom{9}{3}p^3(1-p)^6 p,$$ where $p=0.3$.

For the second problem, it is easier to find first the probability of no prizes. The probability the first results in no prize is $\frac{499500}{500000}$. Given that the first rsults in no prize, the probability the second results in no prize is $\frac{499500}{499999}$. And so on. For the probability of no prize, you will get a product of $10$ terms.

Remark: We get an excellent numerical approximation for the probability of no prize using $\left(\frac{499500}{500000}\right)^{10}$. Even more simply, the probability we win on any buy is about $\frac{500}{500000}$. Since multiple wins are highly unlikely, and $1$ in $1000$ tickets is a winner, the probability of a win is about $10\cdot\frac{1}{1000}$.