Questions: In one lottery game, contestants pick six numbers from 1 through 28 and have to match all six for the big prize (in any order). You'll get twice your money back if you match four out of six numbers. If you buy six tickets, what's the probability of matching four out of six numbers? If you buy six tickets, the probability of matching four out of six numbers is . (Enter your answer as a fraction in lowest terms.)

Solution Steps

To find the probability of matching exactly 4 out of 6 numbers in one ticket, we use the hypergeometric distribution formula:

\[ P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} \]

Where:

• \(N = 28\) (total numbers)
• \(K = 6\) (chosen numbers)
• \(n = 6\) (numbers drawn per ticket)
• \(k = 4\) (matching numbers)

Substituting the values, we have:

\[ P(X = 4) = \frac{\binom{6}{4} \binom{22}{2}}{\binom{28}{6}} \approx 0.0092 \]

The probability of not matching exactly 4 out of 6 numbers in one ticket is given by:

\[ P(\text{not } X = 4) = 1 - P(X = 4) \approx 1 - 0.0092 = 0.9908 \]

The probability of not matching exactly 4 out of 6 numbers in all 6 tickets is:

\[ P(\text{not } X = 4 \text{ in all 6}) = (P(\text{not } X = 4))^6 \approx (0.9908)^6 \approx 0.9461 \]

The probability of matching exactly 4 out of 6 numbers in at least one of the 6 tickets is:

\[ P(X = 4 \text{ in at least one}) = 1 - P(\text{not } X = 4 \text{ in all 6}) \approx 1 - 0.9461 \approx 0.0539 \]

The probability of matching 4 out of 6 numbers in at least one of the 6 tickets can also be expressed as a fraction:

\[ P(X = 4 \text{ in at least one}) \approx \frac{42020}{778929} \]

Final Answer