question_answer

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

\[P(A)=\frac{1}{30}+\frac{4}{5}=\frac{1+24}{30}=\frac{25}{30}\]

\[\therefore \]      \[=\frac{5}{6}\]