A letter is known to have come from either TATANAGAR' or 'CALCUTTA?. On the envelope just two letters ?TA? are visible. What is the probability that the letter has come from |
(i) TATANAGAR? |
(ii) CALCUTTA? |
OR |
If A and B are two independent events such that \[P(\bar{A}\cap B)=\frac{2}{15}\] and \[P(A\cap \bar{B})=\frac{1}{6},\] then find P(A) and P(B). |
Answer:
Let \[{{E}_{1}}\] and \[{{E}_{2}}\] be the events that the letters come from ?TATANAGAR? and ?CALCUTTA?, respectively. Let E be the event that two consecutive letters TA are visible on envelope. Then, \[P({{E}_{1}})=P({{E}_{2}})=\frac{1}{2}\] Also, \[P(E/{{E}_{1}})=\frac{2}{8}\] [\[\because \] in the word ?TATANAGAR?, there are 8 pair of consecutive letters in which TA occurs twice] and \[P(E/{{E}_{2}})=\frac{1}{7}\] [\[\because \] in the word ?CALCUTTA?, there are 7 pair of consecutive letters in which TA occurs once] (i) \[P({{E}_{1}}/E)=\frac{P({{E}_{1}})\cdot P(E/{{E}_{1}})}{P({{E}_{1}})\cdot P(E/{{E}_{1}})+P({{E}_{2}})\cdot P(E/{{E}_{2}})}\] [by Bay?s theorem] \[=\frac{\frac{1}{2}\times \frac{2}{8}}{\frac{1}{2}\times \frac{2}{8}+\frac{1}{2}\times \frac{1}{7}}=\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{7}}\] \[=\frac{\frac{1}{4}}{\frac{7+4}{28}}=\frac{1}{4}\times \frac{28}{11}=\frac{7}{11}\] (ii) \[P({{E}_{2}}/E)=\frac{P({{E}_{2}})\cdot P(E/{{E}_{2}})}{P({{E}_{1}})\cdot P(E/{{E}_{1}})+P({{E}_{2}})\cdot P(E/{{E}_{2}})}\] \[=\frac{\frac{1}{2}\times \frac{1}{7}}{\frac{1}{2}\times \frac{2}{8}+\frac{1}{2}\times \frac{1}{7}}=\frac{\frac{1}{2}\times \frac{1}{7}}{\frac{1}{2}\left[ \frac{1}{4}+\frac{1}{7} \right]}=\frac{\frac{1}{7}}{\frac{7+4}{28}}\] \[=\frac{1}{7}\times \frac{28}{11}=\frac{4}{11}\] OR Given, A and B are two independent events with \[P(\bar{A}\cap B)=\frac{2}{15}\] and \[P(A\cap \bar{B})=\frac{1}{6}.\] We know that, if A and B are independent, then \[\bar{A},\,\,B\] and \[A,\,\,\bar{B}\] are also independent events. Now, \[P(\bar{A}\cap B)=\frac{2}{15}\] \[\Rightarrow \] \[P(B)P(\bar{A})=\frac{2}{15}\] \[\Rightarrow \] \[P(B)[1-P(A)]=\frac{2}{15}\] \[\Rightarrow \] \[P(B)-P(A)P(B)=\frac{2}{15}\] ?(i) and \[P(A\cap \bar{B})=\frac{1}{6}\] \[\Rightarrow \] \[P(A)P(\bar{B})=\frac{1}{6}\] \[\Rightarrow \] \[P(A)[1-P(B)]=\frac{1}{6}\] \[\Rightarrow \] \[P(A)-P(A)P(B)=\frac{1}{6}\] ?(ii) On subtracting (i) from Eq. (ii), we get \[P(A)-P(B)=\frac{1}{6}-\frac{2}{15}=\frac{5-4}{30}=\frac{1}{30}\] \[\Rightarrow \] \[P(A)=\frac{1}{30}+P(B)\] ?(iii) On substituting this value in Eq. (i), we get \[P(B)-\left( \frac{1}{30}+P(B) \right)P(B)=\frac{2}{15}\] Let P(B) = x Then, \[x-\left( \frac{1}{30}+x \right)x=\frac{2}{15}\] \[\Rightarrow \] \[30x=(1+30x)x=4\] \[\Rightarrow \] \[30x-x-30{{x}^{2}}=4\] \[\Rightarrow \] \[30{{x}^{2}}-29x+4=0\] \[\Rightarrow \] \[(6x-1)(5x-4)=0\]\[\Rightarrow \] \[x=\frac{1}{6}\,\,\,\text{or}\,\,x=\frac{4}{5}\] \[\Rightarrow \] \[P(B)=\frac{1}{6}\,\,\text{or}\,\,P(B)=\frac{4}{5}\] [put x = P(B)] Now, if \[P(B)=\frac{1}{6},\] then from Eq. (iii), we get \[P(A)=\frac{1}{30}+\frac{1}{6}=\frac{1+5}{30}=\frac{6}{30}=\frac{1}{5}\] and if \[P(B)=\frac{4}{5},\] then from Eq.(iii), we get \[P(A)=\frac{1}{30}+\frac{4}{5}=\frac{1+24}{30}=\frac{25}{30}\] \[\therefore \] \[=\frac{5}{6}\]
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