Question

A player in the Powerball lottery picks five different integers between \(1\) and \(59\) , inclusive, and a sixth integer between \(1\) and \(39\) , which may duplicate one of the earlier five integers. The player wins the jackpot if the first five numbers picked match the first five number 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 \(200,000\), 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 \(100\) by matching exactly three of the first five and the sixth numbers or 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 \( = \frac{1}{{195249054}}\).

b) The probability that a player wins \(200,000\) is \( = \frac{1}{{5138133}}\).

c) The probability that a player wins \(100\) is \( = \frac{{45}}{{357,590}}\).

d) The probability of winning the prize is \( = 0.9744\).

Step-by-step solution

Given data

A player in the power ball lottery pics \(5\) different integers between \(1\) and \(59\) and a \(6th\) integer between \(1\) and \(39\).

Concept of Probability

If \(S\) is a finite non-empty sample space of equally likely outcomes and \(E\) is an event which is a subset of \(S\).

The probability of \(E\) is \(P(E) = \frac{{\left| E \right|}}{{\left| S \right|}}\).

Calculation for the probability that a player wins the jackpot

a)

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

The number of ways to choose \(5\) numbers from \(59\) numbers is \(C(59,5)\).

The number of ways to choose \(1\) number from \(39\) numbers is \(C(39,1)\).

Therefore, the number possibilities are given as:

\(\begin{aligned}{}C(59,5) \cdot C(39,1) = \frac{{59!}}{{(59 - 5)!5!}} \cdot \frac{{39!}}{{(56 - 1)!!!}}\\C(59,5) \cdot C(39,1) = 195249054\end{aligned}\)

Then, the probability that a player wins the jackpot is \( = \frac{1}{{195249054}}\).

Calculation for the probability that a player wins \(200,000\)

b)

The number of total possibilities is \(C(59,5) \cdot C(39,1) = 195249054\).

The number of ways to choose \(6th\) digit from \(39\) numbers is \(C(39,1) = 39\).

There is only one way to choose the sixth digit, so the remaining \(38\) are wrong to chose it. Then, the probability that a player wins \(200,000\) is \(\frac{{38}}{{195249054}} = \frac{1}{{5138133}}\).

Therefore, the probability that a player wins \(200,000\) is \( = \frac{1}{{5138133}}\).

Calculation for the probability that a player wins \(100\)

c)

The probability of matching exactly \(3\) of the first \(5\) numbers can be determined as:

\(3\)correct numbers are chosen from \(5\) correct numbers and then the two remaining two numbers are chosen from \(54\).

Then, the number of ways can be done is \(C(5,3).C(54,2)\).

Then, the probability that matching three of first five numbers and sixth number correctly is\( = \frac{{C(5,3).C(54,2)}}{{195249054}}\)

The number of ways four of first five numbers are matched and sixth number not matched correctly is\( = C(5,4) \cdot C(54,1) \times 38\)

Then, the probability of winning the prize is given as:

\(\begin{aligned}{}\frac{{C(5,3) \cdot C(54,2) + C(5,4) \cdot C(54,1) \times 38}}{{195249054}} = \frac{{24570}}{{195249054}}\\\frac{{C(5,3) \cdot C(54,2) + C(5,4) \cdot C(54,1) \times 38}}{{195249054}} = \frac{{45}}{{357,590}}\end{aligned}\)

Therefore, the probability of winning the prize is \( = \frac{{45}}{{357,590}}\).

Calculation for the probability of winning the prize

d)

First, we have to find the not winning the prize, that is

The number of ways for zero, one two and sixth number chosen incorrectly is \( = (C(5,0) \cdot C(54,5) + C(5,1) \cdot C(54,4) + C(5,2) \cdot C(54,3)) \times 39\)

Then, the probability of not winning the prize is\(\frac{{\left( {C(5,0) \cdot C(51,5) + C(5,1) \cdot C(51,4) + C(5,2) \cdot C(51,3)} \right)}}{{195249054}} = 0.0256\)

The probability of winning the prize is \(1 - 0.0256 = 0.9744\).

Therefore, the probability of winning the prize is \( = 0.9744\).