<p>On the disk shown above, a player spins the arrow twice. The fraction a/b is formed, where a is the number of the sector where the arrow stops after the first spin and b is the number of the sector where the arrow stops after the second spin. On every spin, each of the numbered sectors has an equal probability of being the sector on which the arrow stops. What is the probability that the fraction a/b is greater than 1?</p>
<p>A. 15/36 B. 16/36 C. 18/36 D. 20/36 E. 21/36</p>
<p>The picture shows a circle with an arrow divided into 6 equal parts (labeled 1, 2, 3...)</p>
<p>Does anyone know how to do this problem?</p>
<p>And any advice for tackling probability/statistics problems in general? they seem to be my weak spot.</p>
<p>Thanks!</p>
<p>Figure out how many possible fractions you could have that are greater than 1...</p>
<p>If b is 1, there are 5 possible a values that would make the fraction greater than one.
If b is 2, there are 4
If b is 3, there are 3
If b is 4, there are 2
If b is 5, theres is 1
If b is 6, there are 0</p>
<p>Add those up, and you have 15 possibilities. A is correct.</p>
<p>15/36</p>
<p>Think of it this way:</p>
<p>There's a 1/6 chance of getting a = 1. So 5/6 of the time our second number (b) will be greater than 1. So 1/6 * 5/6 = 5/36 </p>
<p>There's a 1/6 chance of getting a = 2. So 4/6 of the time, our second number will be greater than 2. So 1/6 * 4/6 = 4/36.</p>
<p>There's a 1/6 chance of getting a = 3. So 3/6 of the time, our second number will be greater than 3. So 1/6 * 3/6 = 3/36.</p>
<p>There's a 1/6 chance of getting a = 4...2/36
There's a 1/6 chance of getting a = 5...1/36
There's a 1/6 chance of getting a =6...0/36</p>
<p>1/6 of the time we get a = b and so a/b = 1 (obv. not greater than 1).</p>
<p>a/b < 1 is 15/36
a= b is 1/6
a/b> 1 is 15/36</p>
<p>does this make sense?</p>