   Chapter 14.4, Problem 19E ### Algebra and Trigonometry (MindTap ...

4th Edition
James Stewart + 2 others
ISBN: 9781305071742

#### Solutions

Chapter
Section ### Algebra and Trigonometry (MindTap ...

4th Edition
James Stewart + 2 others
ISBN: 9781305071742
Textbook Problem

# APPLICATIONSLottery In a 6 / 49 lottery game, a player pays $1 and selects six numbers from 1 to 49 . Any player who has chosen the six winning numbers wins$ 1 , 000 , 000 . Assuming that this is the only way to win, what is the expected value of this game?

To determine

The expected value of this game.

Explanation

Approach:

The expression for probability is given as,

p=n(E)n(S)(1)

Here, n(E) is the number of event and n(S) is the total number of sample.

The compliment of the probability is given as,

p=1p(2)

Here, p is the compliment of the probability and p is the probability.

The expected value is given as,

E=a1p1+a2p2++anpn(3)

Here, a1,a2,,an is the amount of payout and p1,p2,,pn is the probability of payout.

Given:

A player pays $1 and selects six numbers from 1 to 49. If the player choose the six winning numbers then he wins$1,000,000.

Calculation:

The experiment consist of selecting six numbers from the 49 numbers so, the total sample space is C(49,6). And, the way of choosing six winning number from six number is C(6,6).

Calculate the probability of winning.

Substitute C(6,6) for n(E) and C(49,6) for n(S) in equation (1), to calculate the probability.

p1=C(6,6)C(49,6)=113,983,816

Calculate the probability of losing

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