Question

7th Grade Math

Need help with both 6 and 7

Answer 1

Part 6)

case a)
`P(odd)=(1)/(2)=0.5=50\%`

case b)
`P(less\ than\ 3)=(0)/(8)=0=0\%`

Part 7)

case a)
`P(yellow)=(3)/(20)=0.15=15\%`

case b)
`P(green\ and\ red)=(1)/(20)=0.05=5\%`

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=(x)/(y)`

Part 6)

case a) P(odd)

In this problem we have

`x=4`

`y=8`

substitute

`P=(4)/(8)=(1)/(2)=0.5=50\%`

case b) P (less than 3)

In this problem we have

`x=0`

`y=8`

substitute

`P=(0)/(8)=0=0\%`

Part 7)

case a) P(yellow)

In this problem we have

`x=3`

`y=20`

substitute

`P=(3)/(20)=0.15=15\%`

case b) P(green and red)

`P(green)=(5)/(20)`

`P(red)=(4)/(20)`

therefore

`P(green\ and\ red)=(5)/(20)*(4)/(20)=(20)/(400)=(1)/(20)=0.05=5\%`