Dart Game

A game consists of throwing darts onto a target. The target is divided into four sectors as shown in the figure below

Assume that the throws are independent from one another and that the players hit the target on every throw.

1. A player throws a dart. Let p_0 represent the probability of getting 0 point, p_3 the probability of getting 3 points, and p_5 the probability of getting 5 points.
Thus, p_0+p_3+p_5=1. Given p_5=\frac{1}{2}p_3 and p_5=\frac{1}{3}p_0, find the values of p_0,p_3 and p_5.

2. One of the games consists of throwing a maximum of three darts. The player wins the game if she obtains a total (for the three throws) of 8 or more points. If she has a total of 8 or more points after two throws, she does not throw the third dart.
Let W_2 represent the event “the player wins the game in 2 throws”,  W_3 the event “the player wins the game in 3 throws”, and L the event “the player loses the game”. Let p\left (A\right ) represent the probability of an event A.
(a) Show that p\left (W_2\right )=\frac{5}{36} by using a possibility tree
(b) Derive p\left (L\right ) knowing that p\left (W_3\right )=\frac{7}{36}

3. A player plays six games according to the rules given in Question 2. What is the probability that she will win at least one game?

4. The price  for one game is fixed at 2 euros. If the player wins the game in two throws, she receives 5 euros. If she wins in three throws, she receives 3 euros. If she loses, she receives nothing. Let X be the random variable that represents the algebraic gain by the player for one game. The possible values of X are therefore: -2, 1 and 3.
(a) Give the probability distribution of X
(b) Determine the expected value of X. Is the game favorable to the player?
Source: Baccalaureat General, Session Avril 2011, Pondichery, Serie Scientifique, www.ilemaths.net

SOLUTION

1. Find the values of p_0, p_3 and p_5.
We have three unknown variables p_0,p_3, p_5 and three equations
p_0+p_3+p_5=1\quad\quad,\left (1\right )
p_5=\frac{1}{2}p_3\quad\quad\quad\qquad,\left (2\right )
p_5=\frac{1}{3}p_0\quad\quad\quad\qquad,\left (3\right )
Compare Eq. \left (2\right ) and \left (3\right )
\frac{1}{2}p_3=\frac{1}{3}p_0
Multiply both sides by 2
p_3=\frac{2}{3}p_0\quad\quad\quad\qquad,\left (4\right )
Substitute the values of p_3 and p_5 into Eq. \left (1\right )
p_0+\frac{2}{3}p_0+\frac{1}{3}p_0=1
\frac{3p_0+2p_0+p_0}{3}=1
\frac{6p_0}{3}=1
2p_0=1
p_0=\frac{1}{2}
Substitute the value of p_0 into Eq. \left (4\right )
p_3=\frac{2}{3}\times\frac{1}{2}
=\frac{1}{3}
Substitute the value of p_0 into Eq. \left (3\right )
p_5=\frac{1}{3}\times \frac{1}{2}
=\frac{1}{6}

2. (a) Show that p\left (W_2\right )=\frac{5}{36} by using a possibility tree
We will draw the tree symbolically by writing the scores of a W_2 event as follows
5-5, first throw 5 points; second throw 5 points
5-3, first throw 5 points; second throw 3 points
3-5, first throw 3 points; second throw 5 points
p\left (W_2\right )=p_5p_5+p_5p_3+p_3p_5
=\frac{1}{6}\cdot\frac{1}{6}+\frac{1}{6}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{6}
=\frac{1}{36}+\frac{1}{18}+\frac{1}{18}
=\frac{1+2+2}{36}
=\frac{5}{36}

2. (b) Derive p\left (L\right ) knowing that p\left (W_3\right )=\frac{7}{36}
p\left (L\right )+p\left (W_2\right )+p\left (W_3\right )=1
p\left (L\right )+\frac{5}{36}+\frac{7}{36}=1
p\left (L\right )=1-\frac{5}{36}-\frac{7}{36}
=\frac{36-5-7}{36}
=\frac{24}{36}
=\frac{2}{3}

3. A player plays six games according to the rules given in question 2.
What is the probability that she will win at least one game?
Winning at least one game means winning 1,2,3,4,5 or 6 games. In terms of losing that means losing 5,4,3,2,1 or 0 games.
Probability of losing all six games
=\left [p\left (L\right )\right ]^6
=\left (\frac{2}{3}\right )^6
\therefore Probability of winning at least one game
=1-\left (\frac{2}{3}\right )^6
\approx .91

4. (a) Give the probability distribution of X
X=-2 when the player loses the game
p\left (X=-2\right )=p\left (L\right )
=\frac{2}{3}
X=1 when the player wins the game in three throws
p\left (X=1\right )=p\left (W_3\right )
=\frac{7}{36}
X=3 when the player wins the game in two throws
p\left (X=3\right )=p\left (W_2\right )
=\frac{5}{36}

4. (b) Determine the expected value of X. Is the game favorable to the player?
E\left (X\right )=\left (-2\right )p\left (X=-2\right )+1p\left (X=1\right )+3p\left (X=3\right )
=\left (-2\right )\times \frac{2}{3}+1\times \frac{7}{36}+3\times \frac{5}{36}
\approx -.72
The negative expected value indicates that the player is expected to lose money over the long run. The game is not favorable to the player.

Answer
1. p_0=\frac{1}{2};\;p_3=\frac{1}{3};\;p_5=\frac{1}{6}
2. (a) see proof in solution
2. (b) p\left (L\right )=\frac{2}{3}
3. Probability of winning at least one game \approx .91
4. (a) p\left (X=-2\right )=\frac{2}{3};\;p\left (X=1\right )=\frac{7}{36};\;p\left (X=3\right )=\frac{5}{36}
4. (b) E\left (X\right )\approx -.72; game not favorable to player

Advertisements

About mvtrinh

Retired high school math teacher.
This entry was posted in Problem solving and tagged , . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s