# Lottery Probability Questions

In a game of chance, you mark two different numbers from 1 to 25 on a ticket. During the draw, two different numbers are drawn from an urn containing balls with the numbers 1 to 25. If you have marked the drawn numbers (2 correct numbers), you win 250 Euro. If only one of the ticked numbers was drawn (1 correct), then you still win 2.00 euros.

Question 1: The probability of having marked both drawn numbers on a ticket is in

$$(A)(0.003,0.005]\\(B) (0.007,0.009]\\(C) (0.005,0.007]\\(D) (A)-(C) \text{false}.$$

So, my attempt to solve this is the following:

$$P(\text{ picking winning numbers }\\=P(\text{ picking one of the winning numbers on the first draw })\cdot\\P(\text{ picking the other winning number on the second draw })\\=\frac{2}{25}\frac{1}{24}\approx0.0033\implies (A)$$

Question 2: Let A be the event that the ticket with the numbers 2 and 5 has 2 correct numbers. Let B be the event that the ticket with the numbers 2 and 11 has 2 correct numbers. A and B are then

(A) disjoint and independent.

(B) disjoint but not independent.

(C) neither disjoint nor independent.

(D) independent, but not disjoint.

My attempt:

$$A$$ and $$B$$ are disjoint since it's not possible for $$3$$ numbers to be the winning ones. Hence $$P(A\cap B)=0$$.

They are not independent since $$P(A\cap B)=0\neq P(A)P(B)= \frac{2}{600}\frac{2}{600}$$

So $$(B)$$ is the correct answer.

Question 3: The amount in euros that a lottery ticket brings in as winnings on average is in

$$(A) (0.85,0.90].\\ (B) (0.95,1.00].\\ (C) (0.90,0.95].\\ (D) (A)-(C) false.$$

My attempt:

We have that the expectation of the reward is:

$$E(reward)=(250)P(\text{ picking 2 winning numbers } + (2)P(\text{ picking 1 winning number } = (250)\frac{2}{600}+(2)\frac{2}{25}\frac{23}{24}=0.83+0.15=0.98$$

Hence, $$(B)$$ is the right answer.

• You could also explain the answers using binomial coefficients which may help clarify things... having selected the numbers simultaneously rather than in sequence... the answer to (1) having been written as $\frac{\binom{2}{2}\binom{23}{0}}{\binom{25}{2}}$ instead... following from the hypergeometric distribution Commented Aug 26, 2021 at 16:23
• @JMoravitz Yeah I thought I did that one wrong. So I should multiply by $2$ the second member of the sum of $E(reward)$? Commented Aug 26, 2021 at 16:23