Question

A player in the Mega Millions lottery picks five different integers between \(1\) and \(56\) , inclusive, and a sixth integer between \(1\) and \(46\) , which may duplicate one of the earlier five integers. The player wins the jackpot if the first five numbers picked match the first five numbers drawn and the sixth number matches the sixth number drawn.

a) What is the probability that a player wins the jackpot?

b) What is the probability that a player wins \(250,000\), which is the prize for matching the first five numbers, but not the sixth number, drawn?

c) What is the probability that a player wins \(150\) by matching exactly three of the first five numbers and the sixth number or by matching four of the first five numbers but not the sixth number?

d) What is the probability that a player wins a prize, if a prize is given when the player matches at least three of the first five numbers or the last number.

Short answer

Answer

a) The probability that a player wins the jackpot is \( = \frac{1}{{175711536}}\).

b) The probability that a player wins \(250,000\) is \( = \frac{{45}}{{175711536}}\).

c) The probability that a player wins \(150\) is \( = \frac{{24225}}{{175711536}}\).

d) The probability of winning the prize is \( = 0.025\).

Step-by-step solution

Given data

A player in the mega millions lottery picks 5 different integers between 1 and 56 and a 6 th integer between 1 and $46 .$

Concept of Combination

The number of ways to choose \(r\) numbers from \(n\) numbers is \(C(n,r) = \frac{{n!}}{{(n - r)!r!}}\).

Calculation for the probability that a player wins the jackpot

a)

The number of ways to choose \(5\) integers from \(56\) numbers is \(C(56,5)\).

The number of ways to choose \(1\) integers from \(46\) numbers is \(C(56,1)\).

Therefore, the number possibilities are \(C(56,5).C(56,1)\).

\(\begin{aligned}{l}C(56,5) \cdot C(56,1) = \frac{{56!}}{{(56 - 5)!5!}} \cdot \frac{{56!}}{{(56 - 1)!1!}}\\C(56,5) \cdot C(56,1) = 175711536\end{aligned}\)

Then, the probability that a player wins the jackpot is \(\frac{1}{{175711536}}\).

Calculation for the probability that a player wins \(250,000\)

b)

The number of total possibilities is \(C(56,5) \cdot C(56,1) = 175711536\).

The number of ways to choose \(6th\) digit from \(46\) numbers is \(C(46,1)\).

There is only one way to choose the sixth digit, so the remaining \(45\) are wrong to chose it.

Therefore, the probability that a player wins \(250,000\) is \( = \frac{{45}}{{175711536}}\).

Calculation for the probability that a player wins \({\bf{150}}\)

c)

The number of ways, four of the first five numbers are matched and the sixth number is not matched is \(C(5,4).C(51,1) \times 45\).

Then, the probability that a player wins \(150\) is given as:

\(\frac{{C(5,3) \cdot C(51,2) + C(5,4) \cdot C(51,1) \times 45}}{{175711536}} = \frac{{24225}}{{175711536}}\)

Therefore, the probability that a player wins \(150\) is \( = \frac{{24225}}{{175711536}}\).

Calculation for the probability of winning the prize

d)

The number of ways for zero, one two and sixth number chosen incorrectly is \( = (C(5,0) \cdot C(51,5) + C(5,1) \cdot C(51,4) + C(5,2) \cdot C(51,3)) \times 45\).

Then the probability of not winning the prize is \( = \frac{{\mid C(5,0) \cdot C(51,5) + C(5,1) \cdot C(51,4) + C(5,2) \cdot C(51,3)) \times 45}}{{175711536}}\).

The probability of winning the prize is given as:

\(1 - \frac{{|C(5,0) \cdot C(51,5) + C(5,1) \cdot C(51,4) + C(5,2) \cdot C(51,3)| \times 45}}{{175711536}} = 0.025\)

Therefore, the probability of winning the prize is \( = 0.025\).