Probability of Throwing a Dart at the Bull's-Eye

[repugno]

Hello all, can anyone help me on this question? since it doesn’t make any sense to me. Thank you.

A darts player practises throwing a dart at the bull’s-eye on a dart board. Independently for each throw, her probability of hitting the bull’s-eye is 0.2. Let X be the number of throws she makes, up to and including her first success.

(a) Find the probability that she is successful for the first time on her third throw.
(b) Write down an equation for the probability distribution of X
(c) Find the probability that she will have at least 3 failures before her first success.

 

[Hurkyl]

What do you and do you not understand?

P.S. if you are seeking help with a homework problem, you should do it in one of the homework help subforums here.

 

[repugno]

Sorry I posted it here; I figured that I would get a response quicker. I understand that X is the number of throws she makes, and X is therefore a random variable. Also for every throw she has a 20% chance of hitting the target. I need to find out how many times she throws the dart, which I can’t do.

 

[jcsd]

So is it a), b) or c) your stuck on?

for b) find P(X) = f(X)

 

[repugno]

I'm stuck with the whole question, don't know where to start. I thought maybe someone could lend a helping hand.

 

[jcsd]

a)There isa only one permutuation for this which is miss-miss-hit, what is the probailty of this sequence of events happening?

b) generalize your answer from a) which gave you P(3) in order to find P(X)

c) Find the sequences of events where she DOESN'T have 3 failures before she hits, add up their probailties and then 1 minus this number will give you your answer.

 

[repugno]

Thanks. It’s actually so simple. The problem is that this question came under a topic called discrete random variables and so I thought there was some kind of special way I had to use to calculate the probabilities.

There was another one which confused me..

The random variable X has c.d.f F(x) defined by

F(x)=2+x/7; x=1,2,3,4 and 5

Find the probability distrubution for X. Normally I would do the following

x: 1 2 3 4 5
P(X=x):3/7 4/7 5/7 6/7 1

This is obviously wrong because P(X=x)=1

 

[Hurkyl]

F(x) is the cumulative distribution. What does that mean?

 

[repugno]

That is a good question, I think it means running total.

The textbook shows $$F(x_0) = P(X\leq x_0)$$

so $$F(\frac{2 + x}{7}) = P(X\leq \frac{2 + x}{7})$$

The random variable X will take on any values equal to and less than the fraction if I substitute in a value for x, problem is that there are more than one value for x. [?]

Am I getting anywhere?

 

[Hurkyl]

That is a good question, I think it means running total

The textbook shows $$F(x_0) = P(X\leq x_0)$$

That is correct. Recall that the domain is $$\{1, 2, 3, 4, 5\}$$, so, for example, this means

$$F(2)=P(X\leq 2)=P(X = 1)+P(X=2)$$

Can you see where to go from here?

 

[repugno]

Yep, I understand that. But where does the $$F(x) = \frac{2 + x}{7}$$ come in?

 

[Hurkyl]

It tells you the value of $$F(2)$$.

Could you find $$P(X=2)$$ if you knew what $$P(X=1)$$ is?

 

[repugno]

Here is a way which works..
If $$F(x) = \frac{2 + x}{7}$$ only applies for the first value of $$x$$, then $$\frac{2 + 1}{7} = \frac{3}{7}$$ and using good old algebra I can determine the rest of the values because $$P(X=x)=1$$

so, $$\frac{3}{7} + 4x = 1$$

$$x = \frac{1}{7}$$

I don't think this is they way it needs to be done though.

 

[Hurkyl]

None of that really made sense...

I think the first thing you were trying to say is $$F(1) = \frac{3}{7}$$, correct?

You know that $$F(x) = P(X \leq x)$$; do you have a guess as to the value of $$P(X = 1)$$?

 

[repugno]

Originally posted by Hurkyl
None of that really made sense...

I think the first thing you were trying to say is $$F(1) = \frac{3}{7}$$, correct?

Sorry for the late reply.
Yep, that is what I meant.
If $$P(X=1)= \frac{3}{7}$$, then the rest of the values must be $$\frac{1}{7}$$

Thank you very much for your help.

 

[Hurkyl]

That answer is correct, but I'm not entirely sure you got it in the right way...

It looks like you said "Okay, the first one has probability $3/7$, and if we spread the remaining $4/7$ amongst the other 4 values, they must each have probability $1/7$" (which would be incorrect)...


(edit: fixed the tex errors)

 

[repugno]

I think this would probably be the right method.

The c.d.f is $$F(x) = \frac{2 + x}{7}$$

So, $$P(X \leq 1) = \frac{3}{7}$$ and $$P(X \leq 2) = \frac{4}{7}$$

$$P(X \leq 2) - P(X \leq 1) = \frac{1}{7}$$

Which tells me $$P(X=2) = \frac{1}{7}$$

Is that right?

 

[Hurkyl]

Yes, that's exactly right.