88.0k views
4 votes
You toss 4 coins and if you toss all heads you win $9. If you toss anything other than 4 heads you lose $1. What is the Expected Value of the game?

1 Answer

3 votes

The expected value of game is $
(-3)/(8)

Solution:

Given, You toss 4 coins and if you toss all heads you win $9.

When you toss anything other than 4 heads you lose $1

We have to find what is the Expected Value of the game?

The expected value is given as:


\text { expected value }=\text { probability of winning } * \text { amount won }+\text { probability of losing } * \text { amount lost. }

since we toss 4 coins, total outcomes
= 2 * 2 * 2 * 2


\text {Expected value}=(1)/(2 * 2 * 2 * 2) * 9+(2 * 2 * 2 * 2-1)/(2 * 2 * 2 * 2) *(-1)


\begin{array}{l}{\text { Expected value }=(1)/(16) * 9+(16-1)/(16)(-1)} \\\\ {=(9)/(16)+(15)/(16)(-1)} \\\\ {=(9)/(16)-(15)/(16)=(-6)/(16)=(-3)/(8)}\end{array}

Hence, the expected value is
(-3)/(8)

User Crossle Song
by
8.4k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories