Question

Suppose that 10,000 tickets are sold in one lottery and 5000 tickets are sold in another lottery. If a person owns 100 tickets in each lottery, what is the probability that she will win at least one first prize?

Short answer

The probability that she will win at least one first prize is 0.0298.

Step-by-step solution

Given information

10,000 tickets are sold in one lottery, and 5000 tickets are sold in another lottery.

A person owns 100 tickets in each lottery.

Defining events

Let:

\({W_1} = \)The event that she will win first prize in the first lottery

\({W_2} = \) The event that she will win first prize in the second lottery

Computing the required probability

The Additional rule of probability provides the probability of occurrence of either of the events A or B. Mathematically; it is given by:

\({\bf{P}}\left( {{\bf{A}} \cup {\bf{B}}} \right){\bf{ = P}}\left( {\bf{A}} \right){\bf{ + P}}\left( {\bf{B}} \right) - {\bf{P}}\left( {{\bf{A}} \cap {\bf{B}}} \right)\)

Since the events of winning first prize in first and second lotteries are independent; therefore,

\(P\left( {{W_1} \cap {W_2}} \right) = P\left( {{W_1}} \right) \times P\left( {{W_2}} \right)\)

So, using the Additional Rule of Probability, obtaining the probability that she will win at least one first prize as:

\(P\left( {{\rm{at least one first prize}}} \right) = P\left( {{W_1} \cup {W_2}} \right)\)

That is,

\(\begin{aligned}{}P\left( {{W_1} \cup {W_2}} \right) &= P\left( {{W_1}} \right) + P\left( {{W_2}} \right) - P\left( {{W_1} \cap {W_2}} \right)\\ &= \frac{{100}}{{10000}} + \frac{{100}}{{5000}} - P\left( {{W_1}} \right) \times P\left( {{W_2}} \right)\\ &= 0.01 + 0.02 - 0.01 \times 0.02\\ &= 0.0298\end{aligned}\)

Therefore, the required probability is 0.0298.