MATH SOLVE

4 months ago

Q:
# 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

Accepted Solution

A:

Answer: The probability of hitting an odd number three times is [tex]3\dfrac{3}{8}[/tex] times more than the probability of hitting an even number 3 times.Step-by-step explanation:From the given picture , the total total number of sections in the spinner = 5Sections having Odd numbers = 3Sections having Even numbers =2We know that , [tex]\text{Probability}=\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}[/tex]So , probability of hitting an odd number = [tex]\dfrac{3}{5}[/tex]Probability of hitting an even number = [tex]\dfrac{2}{5}[/tex]Since all events are independent of each other , So , probability of hitting an odd number three times = [tex](\dfrac{3}{5})^3=\dfrac{27}{125}[/tex]Probability of hitting an even number three times = [tex](\dfrac{2}{5})^3=\dfrac{8}{125}[/tex]Divide Β [tex]\dfrac{27}{125}[/tex] by [tex]\dfrac{8}{125}[/tex] , we get[tex]\dfrac{27}{125}\div\dfrac{8}{125}\\\\=\dfrac{27}{125}\times\dfrac{125}{8}=\dfrac{27}{8}=3\dfrac{3}{8}[/tex] Hence, the probability of hitting an odd number three times is [tex]3\dfrac{3}{8}[/tex] times more than the probability of hitting an even number 3 times.