+0  
 
0
1317
1
avatar

A board game spinner is divided into four regions labeled $A$, $B$, $C$, and $D$. The probability of the arrow stopping on region $A$ is $\frac{3}{8}$, the probability of it stopping in $B$ is $\frac{1}{4}$, and the probability of it stopping in region $C$ is equal to the probability of it stopping in region $D$. What is the probability of the arrow stopping in region $C$? Express your answer as a common fraction.

 Apr 10, 2015

Best Answer 

 #1
avatar+33661 
+9

It must stop somewhere (i.e. total stopping probability = 1) so, 

 

pA + pB + pC + pD = 1

 

3/8  +  1/4  + 2pC = 1   (because pD = pC)

5/8 + 2pC = 1

2pC = 3/8

pC = 3/16

.

 Apr 10, 2015
 #1
avatar+33661 
+9
Best Answer

It must stop somewhere (i.e. total stopping probability = 1) so, 

 

pA + pB + pC + pD = 1

 

3/8  +  1/4  + 2pC = 1   (because pD = pC)

5/8 + 2pC = 1

2pC = 3/8

pC = 3/16

.

Alan Apr 10, 2015

2 Online Users