Question

a player gets to throw 4 darts at the target shown. Assuming the player will always hit the target, the probability of hitting an odd number three times is *blank* times more than the probability of hitting an even number 3 times

Answer 1

3.375

Explanation:

Answer 2

Answer: The probability of hitting an odd number three times is
`3(3)/(8)` times more than the probability of hitting an even number 3 times.

Explanation:

From the given picture , the total total number of sections in the spinner = 5

Sections having Odd numbers = 3

Sections having Even numbers =2

We know that ,
`\text{Probability}=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}`

So , probability of hitting an odd number =
`(3)/(5)`

Probability of hitting an even number =
`(2)/(5)`

Since all events are independent of each other ,

So , probability of hitting an odd number three times =
`((3)/(5))^3=(27)/(125)`

Probability of hitting an even number three times =
`((2)/(5))^3=(8)/(125)`

Divide
`(27)/(125)` by
`(8)/(125)` , we get

`(27)/(125)/(8)/(125)\\\\=(27)/(125)*(125)/(8)=(27)/(8)=3(3)/(8)`

Hence, the probability of hitting an odd number three times is
`3(3)/(8)` times more than the probability of hitting an even number 3 times.