My Math Forum Probability(Hard)

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 May 31st, 2011, 10:24 PM #1 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 Probability(Hard) 1.a)One letter is selected from each of the names: SIMMS,SMITH,THOMPSON.What is the probability that 2,and only 2 are the same? Answer:0.34 b)A candidate attempts a question to which 5 possible answer have been given,one of them correct.For any question,there is a probability of 1/3 that he knows the correct answer.If he does not know the correct answer he will mark one of the answer at random.He does,in fact,mark the correct answer.What is the probability that he knew the correct answer? Answer:5/7 I need help and explanation for the question.I don even know how to start solve the question.Thank you.
June 1st, 2011, 05:46 AM   #2
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Re: Probability(Hard)

Hello, hoyy1kolko!

Quote:
 a) One letter is selected from each of the names: SIMMS, SMITH, THOMPSON. What is the probability that exactly two letters are the same?[color=beige] . [/color]Answer: 0.34

There is no formula for this problem.
[color=beige]. . [/color]We must make an exhuastive list.

We have:[color=beige]. . [/color]$\begin{Bmatrix}\text{Name A: } & I\,M\,M\,S\,S \\ \\ \\ \\ \text{Name B: } & H\,I\,M\,S\,T \\ \\ \\ \\ \text{Name C: } & H\,M\,N\,O \\ & O\,P\,S\,T \end{Bmatrix}$

And there are five case of doubled letters . . .

$H,\,H: \;\;\begin{Bmatrix}P(A,\text{ any})=&1 \\ \\ \\ P(B,\,H)=&\frac{1}{5} \\ \\ \\ P(C,\,H)=&\frac{1}{8}\end{Bmatrix} \;\;\;P(H,H) \:=\:\frac{1}{40}$

$I,\,I: \;\;\begin{Bmatrix}P(A,\,I)=&\frac{1}{5} \\ \\ \\ P(B,\,I)=&-=&\frac{1}{5} \\ \\ \\ P(C,\text{ any})=&1 \end{Bmatrix} \;\;\;P(I,I) \:=\:\frac{1}{25}$

$T,\,T:\;\;\begin{Bmatrix}P(A,\text{ any})=&1 \\ \\ \\ P(B,\,T)=&\frac{1}{5} \\ \\ \\ P(C,\,T)=&\frac{1}{8}\end{Bmatrix}\;\;\;P(T,T) \:=\:\frac{1}{40}$

$M,\,M:\;\;\begin{Bmatrix}P(A,\,M)=&\frac{2}{5} \\ P(B,\,M)=&\frac{1}{5} \\ P(C,\text{ other})=&\frac{7}{8} \end{Bmatrix} \;\;\;\begin{Bmatrix}P(A,\,M)=&\frac{2}{5} \\ P(B,\text{ other})=&\frac{4}{5} \\ P(C,\,M)=&\frac{1}{8}\end{Bmatrix} \;\;\; \begin{Bmatrix}P(A\text{, other})=&\frac{3}{5} \\ P(B,\,N)=&\frac{1}{5} \\ P(C,\,M)=&\frac{1}{8} \end{Bmatrix}=$

[color=beige]. . . . . . . [/color]$P(M,M) \:=\:\frac{14}{200}\,+\,\frac{8}{200}\,+\,\frac{3} {200} \:=\:\frac{25}{200} \:=\:\frac{1}{8}$

$S,\,S:\;\;\begin{Bmatrix}P(A,\,S)=&\frac{2}{5} \\ \\ P(B,\,S)=&\frac{1}{5} \\ \\ P(C\text{, other})=&\frac{7}{8}\end{Bmatirx} \;\;\;\begin{Bmatrix}P(A,\,S)=&\frac{2}{5} \\ \\ P(B\text{, other})=&\frac{4}{5} \\ \\ P(C,\,S)=&\frac{1}{8} \end{Bmatrix} \;\;\;\begin{Bmatrix}P(A\text{, other})=&\frac{3}{5} \\ \\ P(B,\,S)=&\frac{1}{5} \\ \\ P(C,\,S)=&\frac{1}{8} \end{Bmatrix}=$

[color=beige]. . . . . . . [/color]$P(S,S) \:=\:\frac{14}{200}\,+\,\frac{8}{200}\,+\,\frac{3} {200} \:=\:\frac{25}{200} \:=\:\frac{1}{8}$

$\text{Therefore: }\:P(\text{double letters}) \;=\;\frac{1}{40}\,+\,\frac{1}{25}\,+\,\frac{1}{40 }\,+\,\frac{1}{8}\,+\,\frac{1}{8} \;=\; \frac{68}{200} \;=\;0.34$

June 1st, 2011, 07:00 AM   #3
Math Team

Joined: Dec 2006
From: Lexington, MA

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Re: Probability(Hard)

Hello again, hoyy1kolko!

Quote:
 b) A candidate attempts a question to which 5 possible answer have been given, one of them correct. For any question,there is a probability of 1/3 that he knows the correct answer. If he does not know the correct answer, he will mark one of the answer at random. He does, in fact, mark the correct answer. What is the probability that he knew the correct answer?[color=beige] . [/color]Answer: 5/7

We want: the probability that he knew the answer, given that he marked the correct answer.

$\text{We need Bayes' Formula: }\:P(\text{knew answer}\,|\,\text{marked correct}) \;=\;\frac{P(\text{knew answer}\,\wedge\:\text{marked correct})}{P(\text{marked correct})}\;\;[1]$

[color=beige]. . [/color]$P(\text{knew answer}\,\wedge\:\text{marked correct} \;=\;\frac{1}{3}\,\cdot\,1 \;=\;\frac{1}{3}\;\;[2]$

[color=beige]. . [/color]$P(\text{not know}\,\wedge\,\text{marked correct})\;=\;\frac{2}{3}\,\cdot\,\frac{1}{5} \;=\;\frac{2}{15}\;\;[3]$

$\text{The numerator of [1] is: }\:\frac{1}{3}$

$\text{The denominator of [1] is: }\:\frac{1}{3}\,+\,\frac{1}{15} \:=\:\frac{7}{15}$

$\text{Therefore, [1] becomes: }\:\frac{\frac{1}{3}}{\frac{7}{15}} \:=\:\frac{1}{3}\,\cdot\,\frac{15}{7} \;=\;\frac{5}{7}$

 June 1st, 2011, 08:12 AM #5 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 Re: Probability(Hard) Thank for both answer from soroban and rudrax,it helps me understand better and the explanation are interesting. I still don't understand on soroban's solution on question 1 (a) about probability that 2 letter are the same. And I don't understand how the diagram is formed. I need more explanation on it.Thank you.
June 4th, 2011, 04:55 PM   #6
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Joined: Apr 2007

Posts: 2,140
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Quote:
 Originally Posted by soroban $I,\,I: \;\;\begin{Bmatrix}P(A,\,I)=&\frac{1}{5} \\ \\ \\ P(B,\,I)=&-=&\frac{1}{5} \\ \\ \\ P(C,\text{ any})=&1 \end{Bmatrix} \;\;\;P(I,I) \:=\:\frac{1}{25}$
$P(B,\,I)\,=\,\frac{1}{5}$

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