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

Answer: The probability of hitting an odd number three times is $3\dfrac{3}{8}$ 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 = 5

Sections having Odd numbers = 3

Sections having Even numbers =2

We know that , $\text{Probability}=\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}$

So , probability of hitting an odd number = $\dfrac{3}{5}$

Probability of hitting an even number = $\dfrac{2}{5}$

Since all events are independent of each other ,

So , probability of hitting an odd number three times = $(\dfrac{3}{5})^3=\dfrac{27}{125}$

Probability of hitting an even number three times = $(\dfrac{2}{5})^3=\dfrac{8}{125}$

Divide  $\dfrac{27}{125}$ by $\dfrac{8}{125}$ , we get

$\dfrac{27}{125}\div\dfrac{8}{125}\\\\=\dfrac{27}{125}\times\dfrac{125}{8}=\dfrac{27}{8}=3\dfrac{3}{8}$

Hence, the probability of hitting an odd number three times is $3\dfrac{3}{8}$ times more than the probability of hitting an even number 3 times.

Answer 2

Answer:

3.375

Step-by-step explanation:

(3/5)^3 / (2/5)^3 = 3.375