Question

7th Grade Math Need help with both 6 and 7

Answer 1

Answer:

Part 6)

case a) $P(odd)=\frac{1}{2}=0.5=50\%$

case b) $P(less\ than\ 3)=\frac{0}{8}=0=0\%$

Part 7)

case a) $P(yellow)=\frac{3}{20}=0.15=15\%$

case b) $P(green\ and\ red)=\frac{1}{20}=0.05=5\%$

Step-by-step explanation:

we know that

The probability of an event is the ratio of the size of the event space to the size of the sample space.  

The size of the sample space is the total number of possible outcomes  

The event space is the number of outcomes in the event you are interested in.  

so  

Let

x------> size of the event space

y-----> size of the sample space  

so

$P=\frac{x}{y}$

Part 6)

case a) P(odd)

In this problem we have

$x=4$

$y=8$

substitute

$P=\frac{4}{8}=\frac{1}{2}=0.5=50\%$

case b) P (less than 3)

In this problem we have

$x=0$

$y=8$

substitute

$P=\frac{0}{8}=0=0\%$

Part 7)

case a) P(yellow)

In this problem we have

$x=3$

$y=20$

substitute

$P=\frac{3}{20}=0.15=15\%$

case b) P(green and red)

$P(green)=\frac{5}{20}$

$P(red)=\frac{4}{20}$

therefore

$P(green\ and\ red)=\frac{5}{20}*\frac{4}{20}=\frac{20}{400}=\frac{1}{20}=0.05=5\%$