When buying a ticket an entrant chooses five numbers between 1 and 45 inclusive. To win a prize in a lottery, at least three of the balls drawn must match numbers on one's ticket. To win the jackpot, all five balls drawn must match the numbers on one's ticket.
(a) What is the probability that a given ticket wins a prize?
(b) The population of this country is two million, and on one particular draw each resident buys one ticket, choosing their numbers at random, with each ticket being equiprobable, and all residents choose their tickets independently. What is the probability that there are (i) no jackpot winners? (ii) exactly one jackpot winner? (iii) an even number of jack- pot winners?
I've already attempted this question but I'd like to check and whether my answers are correct.
(a) I used the formula $P(X=k) = \frac{\binom{M}{k}\binom{N-M}{r-k}}{\binom{N}{r}}$ where $N$ is total # of balls, $M$ is # of good balls and $X$ is number of good balls drawn.
Probability of winning would be $P(X≥3) = P(X=3 \cup X=4 \cup X=5) = P(X=3) + P(X=4) + P(X=5) $
= $\frac{\binom{5}{3}\times \binom{40}{2} + \binom{5}{4}\times \binom{40}{1} + \binom{5}{5}\times \binom{40}{0}}{\binom{45}{5}} = \frac{127}{19393} ≈ 0.00655$
(b) (i) Binomial Distribution with parameters $n=M$ & $p=\frac{1}{N}$ with $M=2,000,000$ & $N=1,221,759$ .
$P(X=k) = \binom{M}{k}p^{k}(1-p)^{M-k}$
So $P(X=0) = (1-p)^{M} = (1-(\frac{1}{N}))^{M} = 0.19456... ≈ 0.1946$
(ii) $P(X=k) = \binom{M}{k}p^{k}(1-p)^{M-k}$ again so,
$P(X=1) = \binom{M}{1}p(1-p)^{M-1}= 0.318501... ≈ 0.3185$
(iii) $P(X=2n) = \binom{M}{2n}p^{2n}(1-p)^{M-2n}$
= $\binom{2,000,000}{2n}\left[\frac{1}{1,221,759}\right]^{2n}\left[1-\frac{1}{1,221,759}\right]^{2000000-2n}$ for all $n \in \mathbb Z$
Are my solutions correct? If not could you please point out where I've made mistakes and how I could go about fixing them?
Thank you for taking the time.
P.S. Apologies for the lack of formatting, I just don't know how to do it.
(a) I used the formula $P(X=k) = \frac{\binom{M}{k}\binom{N-M}{r-k}}{\binom{N}{r}}$ where $N$ is total # of balls, $M$ is # of good balls and $X$ is number of good balls drawn.
Probability of winning would be $P(X≥3) = P(X=3 \cup X=4 \cup X=5) = P(X=3) + P(X=4) + P(X=5) $
= $\frac{\binom{5}{3}\times \binom{40}{2} + \binom{5}{4}\times \binom{40}{1} + \binom{5}{5}\times \binom{40}{0}}{\binom{45}{5}} = \frac{127}{19393} ≈ 0.00655$
(b) (i) Binomial Distribution with parameters $n=M$ & $p=\frac{1}{N}$ with $M=2,000,000$ & $N=1,221,759$ .
$P(X=k) = \binom{M}{k}p^{k}(1-p)^{M-k}$
So $P(X=0) = (1-p)^{M} = (1-(\frac{1}{N}))^{M} = 0.19456... ≈ 0.1946$
(ii) $P(X=k) = \binom{M}{k}p^{k}(1-p)^{M-k}$ again so,
$P(X=1) = \binom{M}{1}p(1-p)^{M-1}= 0.318501... ≈ 0.3185$
(iii) $P(X=2n) = \binom{M}{2n}p^{2n}(1-p)^{M-2n}$ = $\sum\limits_{n=0}^{2000000/2}\binom{2,000,000}{2n}\left[\frac{1}{1,221,759}\right]^{2n}\left[1-\frac{1}{1,221,759}\right]^{2000000-2n}$
$\sum\limits_{n=0}^{r/2}\binom{r}{2n}\cdot{a}^{2n}\cdot{b}^{r-2n} = \frac{(a+b)^r +(a-b)^r}{2}$
So then $\sum\limits_{n=0}^{2000000/2}\binom{2,000,000}{2n}\left[\frac{1}{1,221,759}\right]^{2n}\left[1-\frac{1}{1,221,759}\right]^{2000000-2n}$
= $\frac{\left(\frac{1}{1221759}+\frac{1221758}{1221759}\right)^{2000000}+\left(\frac{1}{1221759}-\frac{1221758}{1221759}\right)^{2000000}}{2}$
$≈ 0.5189 $