Let $R$ be the event of drawing the red marble.
Let $E{}^A,E{}^B$, and $E{}^C$ respectively denote the events of selecting the box $A,B$, and $C$.
Total number of marbles $=40$
Number of red marbles = 15
\(\therefore P(R)=\frac{15}{40}=\frac{3}{8}\)
Probability of drawing the red marble from box $A$ is given by P(E_A​|R).
\(\therefore P\left(E_A \mid R\right)=\frac{P\left(E_A \cap R\right)}{P(R)}=\frac{\frac{1}{40}}{\frac{3}{8}}=\frac{1}{15}\)
Probability that the red marble is from box $B$ is P(E_B​|R).
\(\Rightarrow P\left(E_B \mid R\right)=\frac{P\left(E_B \cap R\right)}{P(R)}=\frac{\frac{6}{40}}{\frac{3}{8}}=\frac{2}{5}\)
Probability that the red marble is from box $C$ is P(E_C​|R).
\(\Rightarrow P\left(E_C \mid R\right)=\frac{P\left(E_C \cap R\right)}{P(R)}=\frac{\frac{8}{40}}{\frac{3}{8}}=\frac{8}{15}=0.53\)

\(\begin{array}{|c|c|c|c|c|c|} \hline X & 0 & 1 & 2 & 3 & \text { Total } \\ \hline P(X) & \frac{1}{8} & \frac{3}{8} & \frac{3}{8} & \frac{1}{8} & 1 \\ \hline X P(X) & 0 & \frac{3}{8} & \frac{6}{8} & \frac{3}{8} & \frac{12}{8}=\frac{3}{2} \\ \hline X^{2} P(X) & 0 & \frac{3}{8} & \frac{12}{8} & \frac{9}{8} & 3 \\ \hline \end{array}\)
\(\operatorname{Mean}(\mu)=\Sigma X P(X)=\frac{3}{2}\)
\(\text {Variance }\left(\sigma^{2}\right)-\Sigma X^{2} P(X) \quad \mu^{2}=3 \quad \frac{9}{4}=\begin{aligned}\frac{3}{4} \end{aligned}\)

Let events B₁, B₂, B₃ be the following :
B₁ : the bolt is manufactured by machine A
B₂ : the bolt is manufactured by machine B
B₃ : the bolt is manufactured by machine C
Clearly, B₁, B₂, B₃ are mutually exclusive and exhaustive events and hence, they represent a partition of the sample space.
Let the event E be ‘the bolt is defective’.
The event E occurs with B₁ or with B₂ or with B₃. Given that,
P(B₁) = 25% = 0.25, P (B₂) = 0.35 and P(B₃) = 0.40
Again P(E|B₁) = Probability that the bolt drawn is defective given that it is manufactured by machine A = 5% = 0.05
Similarly, P(E|B₂) = 0.04, P(E|B₃) = 0.02.
Hence, by Bayes' Theorem, we have
\(\mathrm{P}\left(\mathrm{B}_{2} \mid \mathrm{E}\right) =\frac{\mathrm{P}\left(\mathrm{B}_{2}\right) \mathrm{P}\left(\mathrm{E} \mid \mathrm{B}_{2}\right)}{\mathrm{P}\left(\mathrm{B}_{1}\right) \mathrm{P}\left(\mathrm{E} \mid \mathrm{B}_{1}\right)+\mathrm{P}\left(\mathrm{B}_{2}\right) \mathrm{P}\left(\mathrm{E} \mid \mathrm{B}_{2}\right)+\mathrm{P}\left(\mathrm{B}_{3}\right) \mathrm{P}\left(\mathrm{E} \mid \mathrm{B}_{3}\right)} \)
\(=\frac{0.35 \times 0.04}{0.25 \times 0.05+0.35 \times 0.04+0.40 \times 0.02} \)
\(=\frac{0.0140}{0.0345}=\frac{28}{69} \)

S: getting a total of = {(3, 6), (4, 5), (5, 4), (6, 3)}
\(P(S)=\frac{4}{36}=\frac{1}{9} \cdot P(\bar{S})=\frac{8}{9}\)
A can win in 1^st, 3^rd, 5^th, 7^th, ..... throws
\(P(A) =P(S)+[P(\bar{S})]^{2} P(S)+[P(\bar{S})]^{4} P(S)+\cdots \cdot \)
\(=\frac{1}{9}+\left(\frac{8}{9}\right)^{2} \cdot \frac{1}{9}+\left(\frac{8}{9}\right)^{4} \cdot \frac{1}{9}+\ldots \ldots \)
\(=\frac{\frac{1}{9}}{1-\frac{64}{81}}=\frac{9}{17} \quad\left[\begin{array}{l} \text { sum of infinite GP } \\ a+a r+a r^{2}+\ldots=\frac{a}{1-r} \end{array}\right]\)

Let E₁ and E₂ denote the events that first and second group will win. Then,
P(E₁) = 0.6 and P(E₂) = 0.4
Let E be the event of introducing the new product.
Then, \(P\left(\frac{E}{E_1}\right)=0.7 \text { and } P\left(\frac{E}{E_2}\right)=0.3\)
Now, we have to find the probability that new product is introduced by second event.
\(\therefore P\left(\frac{E_2}{E}\right)=\frac{P\left(E_2\right) P\left(\frac{E}{E_2}\right)}{P\left(E_1\right) P\left(\frac{E}{E_1}\right)+P\left(E_2\right) P\left(\frac{E}{E_2}\right)}\)
\(=\frac{0.4 \times 0.3}{0.6 \times 0.7+0.4 \times 0.3}=\frac{0.12}{0.42+0.12}=\frac{0.12}{0.54}\)
= 0.22

Let the events Be:
E₁ : coin is two headed
E₂ : coin is biased(heads 75%)
E₃ : Coin is biased(tails 40%)
and A : coin shows up head
\(P({ E }_{ 1 })=P({ E }_{ 2 })=P({ E }_{ 3 })=\frac { 1 }{ 3 } \)
\(P(A/{ E }_{ 1 })=1,P(A/{ E }_{ 2 })=\frac { 75 }{ 100 } =\frac { 3 }{ 4 } ,\)
By Bayes' theorem
\(P({ E }_{ 1 }​/A)=\frac { P({ E }_{ 1 })P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 })+P({ E }_{ 3 })P(A/{ E }_{ 3 }) } \)
\(=\frac { \frac { 1 }{ 3 } \times 1 }{ \frac { 1 }{ 3 } \times 1+\frac { 1 }{ 3 } \times \frac { 3 }{ 4 } +\frac { 1 }{ 3 } \times \frac { 3 }{ 5 } } \)
\(=\frac{4}{4+3+2}=\frac{4}{9}\)

Let the events E₁ and E₂ be the events when lost card is a diamond and not a diamond respectively.
\(P({ E }_{ 1 })=\frac { 13 }{ 52 } =\frac { 1 }{ 4 } \)and
\(P({ E }_{ 2 })=\frac { 39 }{ 52 } =\frac { 3 }{ 4 } \)
Let A be the event:
"two cards drawn from the remaining pack are diamonds"
\(P(A/{ E }_{ 1 })=\frac { 12\times 11 }{ 51\times 50 } \)
\(P(A/{ E }_{ 2 })=\frac { 13\times 12 }{ 51\times 50 } \)
By Bayes' Theorem
\(P({ E }_{ 1 }/A)=\frac { P({ E }_{ 1 })P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 }) } \)
\(=\frac { \left( \frac { 1 }{ 4 } \right) \left( \frac { 12\times 11 }{ 51\times 50 } \right) }{ \left( \frac { 1 }{ 4 } \right) \left( \frac { 12\times 11 }{ 51\times 50 } \right) +\left( \frac { 3 }{ 4 } \right) \left( \frac { 13\times 12 }{ 51\times 50 } \right) } \)
\(=\frac { 12\times 11 }{ 12\times 11+3\times 13\times 12 } \)
\(=\frac { 132 }{ 132+468 } =\frac { 132 }{ 600 } =\frac { 11 }{ 50 } \)

Let 'p' be the probability of success and 'q' be tha probability of failure.
Then p + q = 1 and p = 2q
Solving, \(p=\frac { 2 }{ 3 } ,q=\frac { 1 }{ 3 } \) Also n = 6.
Required probability = \(P(X\ge 4)\)
= P(4)+P(5)+P(6)
= \(^6C_4\)q²p⁴+\(^6C_5\)q¹p⁵+\(^6C_6\)q⁰p⁶
= \(15{ \left( \frac { 1 }{ 3 } \right) }^{ 2 }{ \left( \frac { 2 }{ 3 } \right) }^{ 4 }+6{ \left( \frac { 1 }{ 3 } \right) }{ \left( \frac { 2 }{ 3 } \right) }^{ 5 }+\left( 1 \right) \left( 1 \right) { \left( \frac { 2 }{ 3 } \right) }^{ 6 }\)
= \({ \left( \frac { 2 }{ 3 } \right) }^{ 4 }\left[ \frac { 15 }{ 9 } +\frac { 4 }{ 3 } +\frac { 4 }{ 9 } \right] =\frac { 31 }{ 9 } { \left( \frac { 2 }{ 3 } \right) }^{ 4 }\)

Let the events be:
E₁ : Selected person is a male.
E₂ : Selected person is afemale
And A : Selected person is grey haired.
\(P({ E }_{ 1 })=P({ E }_{ 2 })=\frac { 1 }{ 2 } \)
\(P(A/{ E }_{ 1 })=\frac { 5 }{ 100 } =\frac { 1 }{ 20 } \)
and \(P(A/{ E }_{ 2 })=\frac { 0.25 }{ 100 } =\frac { 1 }{ 400 } \)
Required probability
= \(\frac { P({ E }_{ 1 }).P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 }) } \)
= \(\frac { \left( \frac { 1 }{ 2 } \right) \left( \frac { 1 }{ 20 } \right) }{ \left( \frac { 1 }{ 2 } \right) \left( \frac { 1 }{ 20 } \right) +\left( \frac { 1 }{ 2 } \right) \left( \frac { 1 }{ 400 } \right) } =\frac { \frac { 1 }{ 20 } }{ \frac { 1 }{ 20 } +\frac { 1 }{ 400 } } =\frac { 20 }{ 21 } \)

Let the events be:
E₁ : Student resides in the hostel
E₂ : Student is a day scholar
and A : Student attains 'A' grade.
\(P({ E }_{ 1 })=\frac { 60 }{ 100 } =\frac { 3 }{ 5 } \)
\(P({ E }_{ 2 })=\frac { 40 }{ 100 } =\frac { 2 }{ 5 } \)
and \(P(A/{ E }_{ 1 })=\frac { 30 }{ 100 } =\frac { 3 }{ 10 } \)
\(P(A/{ E }_{ 2 })=\frac { 20 }{ 100 } =\frac { 2 }{ 10 } \)
By Bayes' Theorem,
\(P({ E }_{ 1 }/A)=\frac { P({ E }_{ 1 })P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 }) } \)
\(=\frac { \left( \frac { 3 }{ 5 } \right) \left( \frac { 3 }{ 10 } \right) }{ \left( \frac { 3 }{ 5 } \right) \left( \frac { 3 }{ 10 } \right) +\left( \frac { 2 }{ 5 } \right) \left( \frac { 2 }{ 10 } \right) } =\frac { 9 }{ 9+4 } =\frac { 9 }{ 13 } \)

Let the events be as below:
E₁ : 2 red balls are transfered from Bag I to Bag II
E₂ : 2 black balls are transferd frm Bag I to Bag II
E₃ : 1 red and 1 black balls are transferd from Bag I to Bag
and A : 1 red ball is drawn from Bag II
\(P({ E }_{ 1 })=\frac { ^{ 3 }{ C }_{ 2 } }{ ^{ 7 }{ C }_{ 2 } } =\frac { 3\times 2 }{ 7\times 6 } =\frac { 1 }{ 7 } \)
\(P({ E }_{ 1 })=\frac { ^{ 4 }{ C }_{ 2 } }{ ^{ 7 }{ C }_{ 2 } } =\frac { 4\times 3 }{ 7\times 6 } =\frac { 2 }{ 7 } \)
\(P({ E }_{ 3 })=\frac { ^{ 3 }{ C }_{ 1 }\times ^{ 4 }{ C }_{ 1 } }{ ^{ 7 }{ C }_{ 2 } } =\frac { 3\times 4 }{ \frac { 7\times 6 }{ 1\times 2 } } =\frac { 3\times 4 }{ 7\times 3 } =\frac { 4 }{ 7 } \)
and \(P(A/{ E }_{ 1 })=\frac { 6 }{ 11 } ,P(A/{ E }_{ 2 })=\frac { 4 }{ 11 } ,P(A/{ E }_{ 3 })=\frac { 5 }{ 11 } \)
By Bayes' Theorem
\(P({ E }_{ 2 }/A)=\frac { P({ E }_{ 2 })P(A/{ E }_{ 2 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 })+P({ E }_{ 3 })P(A/{ E }_{ 3 }) } \)
\(=\frac { \left( \frac { 2 }{ 7 } \right) \left( \frac { 4 }{ 11 } \right) }{ \left( \frac { 1 }{ 7 } \right) \left( \frac { 6 }{ 11 } \right) +\left( \frac { 2 }{ 7 } \right) \left( \frac { 4 }{ 11 } \right) +\left( \frac { 4 }{ 7 } \right) \left( \frac { 5 }{ 11 } \right) } \)
\(=\frac { 8 }{ 6+8+20 } =\frac { 8 }{ 34 } =\frac { 4 }{ 17 } \)

Let the events be:
E₁ : The girl gets 1, 2, 3 and 4 on the dice
E₂ : The girl gets 5 or 6 on the dice.
A : Exactly one head shows up.
\(P({ E }_{ 1 })=\frac { 4 }{ 6 } =\frac { 2 }{ 3 } ,P({ E }_{ 2 })=\frac { 2 }{ 6 } =\frac { 1 }{ 3 } \)
and \(P(A/{ E }_{ 1 })=\frac { 1 }{ 2 } ,P(A/{ E }_{ 2 })=\frac { 3 }{ 8 } \)
By Bayes' Theorem,
\(P({ E }_{ 1 }/A)=\frac { P({ E }_{ 1 })P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 }) } \)
\(=\frac { \frac { 2 }{ 3 } \times \frac { 1 }{ 2 } }{ \frac { 2 }{ 3 } \times \frac { 1 }{ 2 } +\frac { 1 }{ 3 } \times \frac { 3 }{ 8 } } =\frac { \frac { 1 }{ 3 } }{ \frac { 1 }{ 3 } +\frac { 1 }{ 8 } } =\frac { 8 }{ 11 } \)

Let A be the event ‘the number on the card drawn is even and B be the event ‘the number on the card drawn is greater than 3. We have to find P(A|B).
Now, the sample space of the experiment is S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Then A = {2, 4, 6, 8, 10}, B = {4, 5, 6, 7, 8, 9, 10}
and A ∩ B = {4, 6, 8, 10}
\(\text { Also } \quad \mathrm{P}(\mathrm{A})=\frac{5}{10}, \mathrm{P}(\mathrm{B})=\frac{7}{10} \text { and } \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{4}{10}\)
Then \(P(A|B)=\frac { P(A\cap B) }{ P(B) } =\frac { \frac { 4 }{ 10 } }{ \frac { 7 }{ 10 } } =\frac { 4 }{ 7 } \)

\(\text {(i) } E=\{1,3,5\}, F=\{2,3\}, E \cap F=\{3\}\)
\(P(E)=\frac{3}{6}, P(F)=\frac{2}{6}, P(E \cap F)=\frac{1}{6},\)
\(\mathrm{P}(\mathrm{E} / \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}=\frac{1}{6} \div \frac{2}{6}=\frac{1}{2} \)
\(\mathrm{P}(\mathrm{FE})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{E})}=\frac{1}{6} \div \frac{3}{6}=\frac{1}{3} \)
\(\text {(ii) } \mathrm{E}=\{1,3,5\}, \mathrm{G}=\{2,3,4,5\}, \mathrm{E} \cap \mathrm{G}=\{3,5\} \)
\(\mathrm{P}(\mathrm{E})=\frac{3}{6}, \mathrm{P}(\mathrm{G})=\frac{4}{6}, \mathrm{P}(\mathrm{E} \cap \mathrm{G})=\frac{2}{6} \)
\(\mathrm{P}(\mathrm{E} / \mathrm{G})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{G})}{\mathrm{P}(\mathrm{G})}=\frac{2}{6} \div \frac{4}{6}=\frac{2}{4}=\frac{1}{2} \)
\(\mathrm{P}(\mathrm{G} / \mathrm{E})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{G})}{\mathrm{P}(\mathrm{E})}=\frac{2}{6} \div \frac{3}{6}=\frac{2}{3} \)
\(\text {(iii) } \mathrm{E}=\{1,3,5\}, \mathrm{F}=\{2,3\}\}, \mathrm{G}=\{2,3,4,5\} \)
\(\mathrm{E} \cap \mathrm{G}=\{3,5\}, \mathrm{F} \cap \mathrm{G}=\{2,3\}, \)
\((\mathrm{E} \cap \mathrm{F}) \cap \mathrm{G}=\{3\} \)
\(\mathrm{P}(\mathrm{E} \cap \mathrm{G})=\frac{2}{6}, \mathrm{P}(\mathrm{F} \cap \mathrm{G})=\frac{2}{6}, \mathrm{P}[(\mathrm{E} \cap \mathrm{F}) \cap \mathrm{G}]=\frac{1}{6} \)
\( \mathrm{Now} \mathrm{P}(\mathrm{E} \cup \mathrm{F} / \mathrm{G}) \)
\(=\mathrm{P}(\mathrm{E} / \mathrm{G})+\mathrm{P}(\mathrm{F} / \mathrm{G})-\mathrm{P}[(\mathrm{E} \cap \mathrm{F}) / \mathrm{G}] \)
\(=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{G})}{\mathrm{P}(\mathrm{G})}+\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{G})}{\mathrm{P}(\mathrm{G})}-\frac{\mathrm{P}[\mathrm{E} \cap \mathrm{F}) \cap \mathrm{G}]}{\mathrm{P}(\mathrm{G})} \)
\(=\left(\frac{2}{6} \div \frac{4}{6}\right)+\left(\frac{2}{6} \div \frac{4}{6}\right)-\left(\frac{1}{6} \div \frac{4}{6}\right) \)
\(=\frac{2}{4}+\frac{2}{4}-\frac{1}{4}=\frac{3}{4} \)
\(\mathrm{P}(\mathrm{E} \cap \mathrm{F} / \mathrm{G})=\frac{\mathrm{P}[\mathrm{E} \cap \mathrm{F}) \cap \mathrm{G}]}{\mathrm{P}(\mathrm{G})}=\frac{1}{6} \div \frac{4}{6}=\frac{1}{6} \)

\(P(\overset { \_ }{ A } \cup \overset { \_ }{ B } )=1-P(A\cap B)\)
\(\frac { 1 }{ 4 } =1-P(A\cap B)\)
\(P(A\cap B)=1-\frac { 1 }{ 4 } =\frac { 3 }{ 4 } \)
And P(A)P(B) = \(\frac { 1 }{ 2 } \times \frac { 7 }{ 12 } =\frac { 7 }{ 24 } \)
Thus \(P(A\cap B)\neq P(A)P(B)\)
Hence, A and B are not independent.

Here \({ \left( \frac { 1 }{ 2 } +\frac { 1 }{ 2 } \right) }^{ 6 }\)
= \(^{ 6 }{ C }_{ 0 }{ \left( \frac { 1 }{ 2 } \right) }^{ 6 }+^{ 6 }{ C }_{ 1 }{ \left( \frac { 1 }{ 2 } \right) }^{ 6 }+...........+^{ 6 }{ C }_{ 6 }{ \left( \frac { 1 }{ 2 } \right) }^{ 6 }\)
= \({ \left( \frac { 1 }{ 2 } \right) }^{ 6 }\left[ ^{ 6 }{ C }_{ 0 }+^{ 6 }{ C }_{ 1 }+........+^{ 6 }{ C }_{ 6 } \right] \)
= \({ \left( \frac { 1 }{ 2 } \right) }^{ 6 }\left[ ^{ 6 }{ C }_{ 0 }+^{ 6 }{ C }_{ 1 }+^{ 6 }{ C }_{ 2 }+^{ 6 }{ C }_{ 3 }+^{ 6 }{ C }_{ 2 }+^{ 6 }{ C }_{ 1 }+^{ 6 }{ C }_{ 0 } \right] \)
\(^6C_3 \)has the max.value in \(^{ 6 }{ C }_{ 0 },^{ 6 }{ C }_{ 1 },^{ 6 }{ C }_{ 2 }\)and \(^{ 6 }{ C }_{ 3 }\)
Here \(^{ 6 }{ C }_{ 3 }{ \left( \frac { 1 }{ 2 } \right) }^{ 6 }\) is maximum.
Hence, P(X = 3) is most likely outcome.

E : Son on one end = {(s, m, f), (s, f, m), (f, m, s), (m, f, s)}
No.of exhaustive cases = 3! = 6
where s = son, m = mother and f = father 
F : Father in middle = {(m, s ,f), (s, f, m)},
\(E\cap F\)={(m, s, f), (s, f, m)}
\(P(E\cap F)=\frac { 2 }{ 6 } =\frac { 1 }{ 3 } \)
and \(P(F)=\frac { 2 }{ 6 } =\frac { 1 }{ 3 } \)
\(P(E/F)=\frac { P(E\cap F) }{ P(F) } =\frac { \frac { 1 }{ 3 } }{ \frac { 1 }{ 3 } } =1\)

We have:
P(A) = 0.3 and P(B) = 0.4
(i) Since A and B are independent,
\(P(A\cap B)=P(A)P(B)\)
= (0.3) (0.4) = 0.12
(ii) \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)
= 0.3 + 0.4 - 0.12 = 0.7 - 0.12 = 0.58
(iii) \(P(A/B)=\frac { P(A\cap B) }{ P(B) } =\frac { 0.12 }{ 0.4 } \)
= \(\frac { 12 }{ 100 } \times \frac { 10 }{ 4 } \)
= \(\frac { 12 }{ 10 } =\frac { 3 }{ 10 } \)
(iv) \(P(B/A)=\frac { P(A\cap B) }{ P(A) } =\frac { 0.12 }{ 0.4 } =\frac { 12 }{ 100 } \times \frac { 10 }{ 3 } =\frac { 12 }{ 10 } =\frac { 2 }{ 5 } \)

The sample space has 216 outcomes.
\(\text { Now } \quad \mathrm{A}=\left\{\begin{array}{lllll} (1,1,4) & (1,2,4) & \ldots & (1,6,4) & (2,1,4) & (2,2,4) & \ldots (2,6,4) \\ (3,1,4) & (3,2,4) & \ldots &(3,6,4) & (4,1,4) & (4,2,4) & \ldots(4,6,4) \\ (5,1,4) & (5,2,4) & \ldots & (5,6,4) & (6,1,4) & (6,2,4) & \ldots(6,6,4) \end{array}\right\}\)
B = {(6,5,1), (6,5,2), (6,5,3), (6,5,4), (6,5,5), (6,5,6)} and A ∩ B = {(6,5,4)}.
\(\text { Now }P(B)=\frac{6}{216} \text { and } P(A \cap B)=\frac{1}{216} \)
Then  \(P(A|B)=\frac { P(A\cap B) }{ P(B) } =\frac { \frac { 1 }{ 216 } }{ \frac { 6 }{ 216 } } =\frac { 1 }{ 6 }\)