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
