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Every time I use a piece of scrap paper, I crumple it up and try to shoot it inside the recycling bin across the room. I'm pretty good at it: If I shoot 5 pieces of paper at the recycling bin, at least one of them will make it inside the recycling bin with probability 211/243. If I shoot 6 pieces of paper at the recycling bin, what's the probability at least two of them make it inside the recycling bin?

(Assume that the probability I make any given shot is uniform across every shot. Leave your answer in exact form; there should be no decimals in your answer.)

 Dec 13, 2019
 #1
avatar+107087 
+1

This question was answered a short time back so go look for the answer.

(I suspect it is the same person asking this time as last anyway so it should be easy to find.  Of course it would be easier if were a member.)

 Dec 13, 2019
 #2
avatar+1832 
+1

The solution(s) presented in the prior posts have one or more errors; most notably an incorrect value for (p) –the probability of a single (successful) event

 

To solve: Find (p), the probability of a single event.

 

Statement:  If I shoot 5 pieces of paper at the recycling bin, at least one of them will make it inside the recycling bin with probability \(\frac {211}{243}\)

Equivalent statistical statement:

\(\Pr(X\geq 1)= 1- \left(\sum_ \limits {i=0}^{1} \binom{5}{i} p^{i}(1-p)^{5-i}\right)=\frac {211}{243}\)

\(\text {Equivalent statement: $Pr(X\geq 1)= \Pr(X > 0) = (\frac {211}{243})$  |  Binomial events are discreet integers. }\)

\(\text {Complementary statement $Pr(X < 1) = \Pr(X = 0) =  \frac {32}{243}$ | Probability of zero (0) successes in five attempts. } \)

\(\Pr(X = 0) =  \binom{5}{0} p^{5}(1-p)^{0}=\frac {32}{243} \)

\(\rightarrow p^5=\frac {32}{243}  \text { |  Solve complementary statement for (p)} \)

\( p = \sqrt [\leftroot{-1}\uproot{1}5]{\frac {32}{243}}\rightarrow p = \frac {2}{3} \text { |  Probability of failure on a single throw}\\ \)

\(1-\frac {2}{3} =\frac {1}{3}  \text { |  Complement of failure = success }\\ \)

\(\text {Use probability of success on a single throw  $(\frac {1}{3})$  to solve:} \\ \) 

If I shoot 6 pieces of paper at the recycling bin, what's the probability at least two of them make it inside the recycling bin

Equivalent statistical statement:

\(\Pr(X\geq 2)= 1- \left(\sum_ \limits {i=0}^{1} \binom{6}{i}(\frac {1}{3})^{i}(1-\frac {1}{3})^{6-i}\right)=64.88\% \)

 

\({}\)

 

GA

 Dec 14, 2019
edited by GingerAle  Dec 14, 2019
 #3
avatar+107087 
+1

Ahr yes, I did make that error.  Thanks Ginger.  laugh

Melody  Dec 14, 2019
 #4
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+1

Probability of getting at least one piece inside the bin out of 5 throws is \(\Pr(X\geq 1)= 1- \left(\sum_ \limits {i=0}^{0} \binom{5}{i} p^{i}(1-p)^{5-i}\right)= 1-(1-p)^{5}=\frac {211}{243}\neq1- \left(\sum_ \limits {i=0}^{1} \binom{5}{i} p^{i}(1-p)^{5-i}\right)\)

 

when p is the probability of success on a single throw.

Guest Dec 15, 2019

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