Need help! probability

10. When Ayla plays darts the chances that she hits a bulls eye is 0.5.


[2] a) Draw a tree diagram to represent this problem when 3 darts are thrown.










[1] b) What are the chances that three darts fired in succession will all hit bulls eye?
= 0.5/3 = 0.16

[1] c) What is the probability that none will hit?

[2] d) What is the probability that at least one will hit?

[2] e) What is the probability that 2 will hit?
0.5 is the probability that it will hit a bullseye.
0.5/2 = 0.25 is the probability that 2 will hit.

I'm not sure how to answer this question or if I am doing the right thing.

Thanks a lot!

kay_wink08 said:

10. When Ayla plays darts the chances that she hits a bulls eye is 0.5.
[1] b) What are the chances that three darts fired in succession will all hit bulls eye?
= 0.5/3 = 0.16

Click to expand...

I will get you started.
Probability that all three hit is \(\displaystyle (0.5)^3~.\)

[1] b) What are the chances that three darts fired in succession will all hit bulls eye?
= (0.5)³

[1] c) What is the probability that none will hit?
(0.5)⁰ =

[2] d) What is the probability that at least one will hit?
Opposite of none = 1 - none
1 - none = 1 - 5⁰

[2] e) What is the probability that 2 will hit?
(0.5)² =

Thanks!!!

Am I on the right path now??

kay_wink08 said:

[1] b) What are the chances that three darts fired in succession will all hit bulls eye?
= (0.5)³

[1] c) What is the probability that none will hit?
(0.5)⁰ =

[2] d) What is the probability that at least one will hit?
Opposite of none = 1 - none
1 - none = 1 - 5⁰

[2] e) What is the probability that 2 will hit?
(0.5)² =
Am I on the right path now??

Click to expand...

no!

None is also \(\displaystyle (0.5)^3\) WHY?

Two hit is \(\displaystyle \binom{3}{2}(0.5)^3\) WHY?

Hello, kay_wink08!

10. When Ayla plays darts the chances that she hits a bulls eye is 0.5.

a) Draw a tree diagram to represent this problem when 3 darts are thrown.
. . Obviously, you didn't do this.

b) What are the chances that three darts fired in succession will all hit bulls eye?

c) What is the probability that none will hit?

d) What is the probability that at least one will hit?

e) What is the probability that 2 will hit?

Click to expand...

We are given: .\(\displaystyle P(\text{hit}) = \frac{1}{2},\;P(\text{miss}) = \frac{1}{2}\)

I won't draw the tree diagram, but I will list the 8 possible outcomes.


. . \(\displaystyle \begin{array}{cccc}\text{Outcome} & \text{Probability} \\ \hline \\ HHH & (\frac{1}{2})(\frac{1}{2})(\frac{1}{2}) \:=\:\frac{1}{8} \\ \\
HHM & (\frac{1}{2})(\frac{1}{2})(\frac{1}{2}) \:=\:\frac{1}{8} \\ \\
HMH & (\frac{1}{2})(\frac{1}{2})(\frac{1}{2}) \:=\:\frac{1}{8} \\ \\
HMM & (\frac{1}{2})(\frac{1}{2})(\frac{1}{2}) \:=\:\frac{1}{8} \\ \\
MHH & (\frac{1}{2})(\frac{1}{2})(\frac{1}{2}) \:=\:\frac{1}{8} \\ \\
MHM & (\frac{1}{2})(\frac{1}{2})(\frac{1}{2}) \:=\:\frac{1}{8} \\ \\
MMH & (\frac{1}{2})(\frac{1}{2})(\frac{1}{2}) \:=\:\frac{1}{8} \\ \\
MMM & (\frac{1}{2})(\frac{1}{2})(\frac{1}{2}) \:=\:\frac{1}{8} \end{array}\)


Can you answer the questions now?