My Math Forum need a help with 2 probabilities

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 April 4th, 2011, 10:26 AM #1 Member   Joined: Jul 2010 Posts: 95 Thanks: 0 need a help with 2 probabilities Code: 1.The pedestrian can cross a street within 3 seconds.The pedestrian start walk then he sees that in 3 seconds period no car cross a street.Probability that within 1 second a car will cross a street is 0.54.what probability that the pedestrian will wait 2 seconds until he gets a chance to cross a street Code: 2.Hunter A has probability to shoot successfully a target 0.56.Hunter B has probability to shoot successfully a target 0.65.Independently they are shooting into target until the first successful hit.What probability that for the first hunter will need more ammunition than for the second
April 4th, 2011, 11:22 AM   #2
Math Team

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From: Lexington, MA

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Re: need a help with 2 probabilities

Hello, safyras!

The second problem is not stated clearly.
Some clarification is needed.

Quote:
 2. Hunter A can hit a target with probability 0.56. Hunter B can hit the target with probability 0.65. Independently they shoot at a target until the first hit. What probability that for the first hunter will need more ammunition than the second?

We must assume a number of rules . . . which is always dangerous.

Do the two hunters fire at the target at the same time, firing the same number of times?
[color=beige]. . [/color](If they do not fire together, do they take turns? . . . A-B-A-B- ... )

The problem says "... until the first hit".
[color=beige]. . [/color]This means that they stop, right?
Then why would a hunter need "more ammunition"?

April 4th, 2011, 11:53 AM   #3
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Re: need a help with 2 probabilities

Quote:
 The problem says "... until the first hit". . . This means that they stop, right? Then why would a hunter need "more ammunition"?
Yes , they stop .
Quote:
 Then why would a hunter need "more ammunition"?
Two hunters shooting at the same time.What probability that hunter A will use more bullets than hunter B to hit a target.

April 4th, 2011, 12:47 PM   #4
Math Team

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Re: need a help with 2 probabilities

Hello again, safyras!

Quote:
 Two hunters shooting at the same time. What probability that hunter A will use more bullets than hunter B to hit a target?

That still doesn't clear up the mystery.

[color=beige]. . [/color]$\begin{array}{c|cc} & \text{A} & \text{B} \\ \\ \\ \hline \text{Round 1} & \text{miss} & \text{miss} \\ \\ \\ \text{Round 2} & \text{miss} & \text{miss} \\ \\ \\ \text{Round 3} & \text{miss} & \text{ miss} \\ \\ \\ \text{Round 4} & \text{miss} & \text{hit} \end{array}$

and they stop . . . So hunter A will not get a chance to use "more bullets".

Could the question be: "What is the probability that Hunter B gets the first hit?"

April 4th, 2011, 01:06 PM   #5
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Re: need a help with 2 probabilities

Quote:
 Could the question be: "What is the probability that Hunter B gets the first hit?"
okey.How to solve it?

 April 5th, 2011, 02:47 AM #6 Member   Joined: Jul 2010 Posts: 95 Thanks: 0 Re: need a help with 2 probabilities Code:  1.The pedestrian can cross a street within 3 seconds.The pedestrian start walk then he sees that in 3 seconds period no car cross a street.Probability that within 1 second a car will cross a street is 0.54.what probability that the pedestrian will wait 2 seconds until he gets a chance to cross a street could i solve the first in this way: $(1-P)^5 \cdot P + (1-P)^4 \cdot P + (1-P)^3 \cdot P$ where P=0.54
April 5th, 2011, 03:21 AM   #7
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Re: need a help with 2 probabilities

Quote:
 Could the question be: "What is the probability that Hunter B gets the first hit?"
so could i solve it like that
$P=0.56 *(1-0.65)\ ?$

 April 5th, 2011, 07:25 AM #8 Math Team   Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408 Re: need a help with 2 probabilities Hello, safyras! Thanks for the clarification . . . I think I have the answer. I had to use baby-talk to get the picture. $\text{Hunter A hits the target with probability } p=0.56\text{ and misses with probability }0.44$ [color=beige]. . [/color]$\text{Let: }\:\begin{Bmatrix}P(\text{A hits})=&p \\ \\ \\ P(\text{A miss})=&1-p \end{Bmatrix}=$ $\text{Hunter B hits the target with probability }q=0.65\text{ and misses with probability }0.35$ [color=beige]. . [/color]$\text{Let: }\;\begin{Bmatrix} P(\text{B hits})=&q \\ \\ \\ P(\text{B miss})=&1-q \end{Bmatrix}=$ $[1]\;B\text{ wins in the 1st round: }\;\;\;\begin{array}{c|c|c} \text{Round} & A & B \\ \hline 1 & \text{miss} & \text{hit}\end{array}$ [color=beige]. . [/color]$P(\text{B wins, 1st round}) \;=\;(1-p)q$ $[2]\;B\text{ wins in the 2nd round: }\;\;\;\begin{array}{c|c|c} \text{Round} & A & B \\ \hline 1 & \text{miss} & \text{miss} \\ 2 & \text{miss} & \text{hit}\end{array}$ [color=beige]. . [/color]$P(\text{B wins, 2nd round}) \;=\;(1-p)^2(1-q)q$ $[3]\;B\text{ wins in the 3rd round: }\;\;\;\begin{array}{c|c|c} \text{Round} & A & B \\ \hline 1 & \text{miss} & \text{miss} \\ 2 & \text{miss} & \text{miss} \\ 3 & \text{miss} & \text{hit} \end{array}$ [color=beige]. . [/color]$P(\text{B wins, 3rd round}) \;=\;(1-p)^3(1-q)^2q$ $[4]\;B\text{ wins in the 4th round: }\;\;\;\begin{array}{c|c|c} \text{Round} & A & B \\ \hline 1 & \text{miss} & \text{miss} \\ 2 & \text{miss} & \text{miss} \\ 3 & \text{miss} & \text{miss} \\ 4 & \text{miss} & \text{hit} \end{array}$ [color=beige]. . [/color]$P(\text{B wins, 4th round}) \;=\;(1-p)^4(1-q)^3q$ $\text{I finally see the pattern . . .}$ $P(\text{B wins}) \;=\;(1-p)q\,+\,(1-p)^2(1-q)q \,+\,(1-p)^3(1-q)^2q \,+\,(1-p)^4(1-q)^3q\,+\, \cdots$ [color=beige]. . . . . . . . . [/color]$=\;(1-p)q\, \underbrace{\bigg[\,1\,+\,(1-p)(1-q)\,+\,(1-p)^2(1-q)^2\,+\,(1-p)^3(1-q)^3\,+\,\cdots\, \bigg]}_{\text{Infinite series}}$ $\text{The infinite series has the sum: }\;\frac{1}{1-(1-p)(1-q)} \:=\:\frac{1}{p\,+\,q\,-\,pq}$ [color=beige]. . [/color]$\text{Hence: }\:P(\text{B wins}) \;=\;\frac{(1-p)q}{p\,+\,q\,pq}$ $\text{Therefore: }\;P(\text{B wins}) \;=\;\frac{(0.44)(0.65)}{0.56\,+\,0.65\,-\,(0.56)(0.65)} \;=\;\frac{0.286}{0.847} \;=\;\frac{143}{423} \;\approx\;33.8\%$ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ $\text{That answer bothered me . . . do you see why?}$ $\text{Since Hunter B is the better shot,}$ [color=beige]. . [/color]$\text{it seems unlikely that he wins only }one\text{-}third\text{ of the time.}$ $\text{So, I did a complete analysis of the problem.}$ $\text{This is what I found:}$ [color=beige]. . [/color]$\begin{array}{c} \text{A wins 23.2\% of the time.} \\ \text{B wins 33.8\% of the time.} \\ \text{They tie 43.0\% of the time.}\;\; \end{array}$ $\text{Quite an eye-opener!}$
 April 5th, 2011, 11:48 AM #9 Member   Joined: Jul 2010 Posts: 95 Thanks: 0 Re: need a help with 2 probabilities Thanks for solution,soroban.How to solve the first problem?

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