Probability throwing darts

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barabbas

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Would be thankful if anyone could help me out here.

Person A, B and C throw darts. A starts, then B then C. When C has thrown his dart, A gets a new chance, then B and so on. First one to hit the target wins and the contest is over. Every time A throws the probability of hitting the target is 1/4, B 1/3 and C 1/2. What is the probability that A wins?
 
barabbas said:
Would be thankful if anyone could help me out here.

Person A, B and C throw darts. A starts, then B then C. When C has thrown his dart, A gets a new chance, then B and so on. First one to hit the target wins and the contest is over. Every time A throws the probability of hitting the target is 1/4, B 1/3 and C 1/2. What is the probability that A wins?
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Please share your work/thoughts about this assignment.
 
barabbas said:
Person A, B and C throw darts. A starts, then B then C. When C has thrown his dart, A gets a new chance, then B and so on. First one to hit the target wins and the contest is over. Every time A throws the probability of hitting the target is 1/4, B 1/3 and C 1/2. What is the probability that A wins?
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Here in something for you to consider:
if \(A\) wins then it will be on throws \(1,~4,~7,~10~ect.\cdots\)
 
Suppose that \(X\) is the number of his term on which \(A\) wins.
Then \(\mathscr{P}(X=1)=\frac{1}{4}\) ie he wins his first tern
\(\mathscr{P}(X=2)=\left(\frac{1}{4}\right)^2\) ie he wins on his second tern. Can you explain?
\(\mathscr{P}(X=n)=\left(\frac{1}{4}\right)^n\) ie he wins on his \(n^{th}\) tern. Can you explain?
Can you evaluuate \(\sum\limits_{k = 1}^\infty {{{\left( {\frac{1}{4}} \right)}^k}} \) Why does that answer the question?
 
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pka said:
Suppose that \(X\) is the number of his term on which \(A\) wins.
Then \(\mathscr{P}(X=1)=\frac{1}{4}\) ie he wins his first tern
\(\mathscr{P}(X=2)=\left(\frac{1}{4}\right)^2\) ie he wins on his second tern. Can you explain?
\(\mathscr{P}(X=n)=\left(\frac{1}{4}\right)^n\) ie he wins on his \(n^{th}\) tern. Can you explain?
Can you evaluuate \(\sum\limits_{k = 1}^\infty {{{\left( {\frac{1}{4}} \right)}^k}} \) Why does that answer the question?
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pka, I am sorry but I am not following what you are saying here. Can you please explain? Here is why I think that you are wrong. Suppose that B always hits the target thenP(x=2) = P(x=3) = ... =0.
Using the numbers given, wouldn't P(X=2) = (3/4)*(2/3)*(1/2)*(1/4)=1/16?
 
Jomo said:
pka, I am sorry but I am not following what you are saying here. Can you please explain? Here is why I think that you are wrong. Suppose that B always hits the target thenP(x=2) = P(x=3) = ... =0.
Using the numbers given, wouldn't P(X=2) = (3/4)*(2/3)*(1/2)*(1/4)=1/16?
Click to expand...
Happy to reply. If no one scores it would be \(\mathscr{P}(\overline{A}\overline{B}\overline{C})=\left(\frac{3}{4}\right)\left(\frac{2}{3}\right)\left(\frac{1}{2}\right)=\frac{1}{4}\)​
So for \(A\) to win when he throws his dart the sixth time it means the all other three did not win on their first five times and \(A\) wins his sixth throw.​
Now that is \(\left(\frac{1}{4}\right)^5\cdot\frac{1}{4}=\left(\frac{1}{4}\right)^6\)​
Thus \(A\) wins on his \(N\) throw with the probability of \(\left(\frac{1}{4}\right)^N\).​
 
pka said:
Happy to reply. If on one scores it would be \(\mathscr{P}(\overline{A}\overline{B}\overline{C})=\left(\frac{3}{4}\right)\left(\frac{2}{3}\right)\left(\frac{1}{2}\right)=\frac{1}{4}\)​
So for \(A\) to win when he throws his dart the sixth time it means the all other three did not win on their first five times and \(A\) wins his sixth throw.​
Now that is \(\left(\frac{1}{4}\right)^5\cdot\frac{1}{4}=\left(\frac{1}{4}\right)^6\)​
Thus \(A\) wins on his \(N\) throw with the probability of \(\left(\frac{1}{4}\right)^N\).​
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Of course. Thanks for the reply!
 
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