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earnstyle [38]
3 years ago
10

In a lottery game, a player picks six numbers from 1 to 27. If the player matches all six numbers, they win 40,000 dollars. Othe

rwise, they lose $1. Hint: There are 296010 possible ways of choosing six numbers, and you want just the one winning combination.
What is the expected value of this game?
Mathematics
1 answer:
Alexus [3.1K]3 years ago
7 0

Answer:

We conclude that  expected value of this game is -0.865$.

Step-by-step explanation:

We know that in a lottery game, a player picks six numbers from 1 to 27.

We know that

C_6^{27}=296010

As there is only one advantageous combination, we conclude that the number of non-winning combinations is 296009.

He can win 40,000 dollars.

We calculate:

E(X)=\frac{1}{296010}\cdot 40000\$- \frac{296009}{296010}\cdot 1\$\\\\E(X)=\frac{40000-296009}{296010}\, \$\\\\E(X)=-0.865\, \$

We conclude that  expected value of this game is -0.865$.

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