Q.In a certain lottery 10,000 tickets are sold and ten equal prizes are awarded. What is the probability of not getting a prize if you buy

Concept: Combinations Probability — we count favourable outcomes as ways to pick tickets that don’t win, divided by total ways to pick your tickets.

1. One ticket Total ways to choose 1 ticket: $\binom{10000}{1} = 10000$. Ways to choose a non‑prize ticket: $\binom{9990}{1} = 9990$. Probability $= \frac{9990}{10000} = 0.999$.
2. Two tickets Total ways: $\binom{10000}{2}$. Ways to pick 2 non‑prize tickets: $\binom{9990}{2}$. Probability $= \frac{\binom{9990}{2}}{\binom{10000}{2}} = \frac{9990 \times 9989}{10000 \times 9999}$.
3. Ten tickets …

Concept and Intuition

This is a classic combinations probability problem. The key idea: when you buy tickets, you are selecting a subset of tickets from the total pool. The lottery awards prizes to 10 specific tickets (the winning ones). You want the probability that none of your chosen tickets are among those 10 winners.

Because the tickets are distinct and the prizes are awarded to specific tickets, the number of ways to choose your tickets is given by combinations ($\binom{n}{r}$). The probability of "not getting a prize" is simply the number of ways to pick your tickets entirely from the non-winning pool divided by the total number of ways to pick your tickets from all tickets.

Let's define:

• Total tickets: $N = 10000$
• Winning tickets: $W = 10$
• Non-winning tickets: $L = N - W = 9990$
• Tickets you buy: $n$ (where $n = 1, 2, \text{ or } 10$)

The probability of no prize is:

Now we compute for each case.

(a) One ticket

You buy just 1 ticket. The only way to not get a prize is if that single ticket is one of the 9990 losers.

(b) Two tickets

You buy 2 tickets. You need both to be from the losing pool.

Let's simplify without computing huge numbers. Write out the combinations:

The $\frac{1}{2}$ cancels, so:

$P = \frac{9990 \times 9989}{10000 \times 9999}$ …