A black and a red dice are rolled.(a) Find the conditional probability of obtaining a sum greater than 9 , given that the black die resulted in a 5 . (b) Find the conditional probability of obtaining the sum 8 , given that the red die resulted in a number less than 4 .
A black and a red dice are rolled.(a) Find the conditional probability of obtaining a sum greater than 9 , given that the black die resulted in a 5 . (b) Find the conditional probability of obtaining the sum 8 , given that the red die resulted in a number less than 4 .

Solution:
Assuming: B denote black coloured die and R denote red coloured die.

Therefore, the sample space for the specified experiment will be as follows:

\mathrm{S}=\left{\begin{array}{l}(\mathrm{B} 1, \mathrm{R} 1),(\mathrm{B} 1, \mathrm{R} 2),(\mathrm{B} 1, \mathrm{R} 3),(\mathrm{B} 1, \mathrm{R} 4),(\mathrm{B} 1, \mathrm{R} 5),(\mathrm{B} 1, \mathrm{R} 6), \ (\mathrm{B} 2, \mathrm{R} 1),(\mathrm{B} 2, \mathrm{R} 2),(\mathrm{B} 2, \mathrm{R} 3),(\mathrm{B} 2, \mathrm{R} 4),(\mathrm{B} 2, \mathrm{R} 5),(\mathrm{B} 2, \mathrm{R} 6), \ (\mathrm{B} 3, \mathrm{R} 1),(\mathrm{B} 3, \mathrm{R} 2),(\mathrm{B} 3, \mathrm{R} 3),(\mathrm{B} 3, \mathrm{R} 4),(\mathrm{B} 3, \mathrm{R} 5),(\mathrm{B} 3, \mathrm{R} 6), \ (\mathrm{B} 4, \mathrm{R} 1),(\mathrm{B} 4, \mathrm{R} 2),(\mathrm{B} 4, \mathrm{R} 3),(\mathrm{B} 4, \mathrm{R} 4),(\mathrm{B} 4, \mathrm{R} 5),(\mathrm{B} 4, \mathrm{R} 6), \ (\mathrm{B} 5, \mathrm{R} 1),(\mathrm{B} 5, \mathrm{R} 2),(\mathrm{B} 5, \mathrm{R} 3),(\mathrm{B} 5, \mathrm{R} 4),(\mathrm{B} 5, \mathrm{R} 5),(\mathrm{B} 5, \mathrm{R} 6), \ (\mathrm{B} 6, \mathrm{R} 1),(\mathrm{B} 6, \mathrm{R} 2),(\mathrm{B} 6, \mathrm{R} 3),(\mathrm{B} 6, \mathrm{R} 4),(\mathrm{B} 6, \mathrm{R} 5),(\mathrm{B} 6, \mathrm{R} 6)\end{array}\right}

(a) Let \mathrm{A} be the event of ‘obtaining a sum greater than 9 ‘ and \mathrm{B} be the event of ‘getting a 5 on black die’.

Then, \mathrm{A}={(\mathrm{B} 4, \mathrm{R} 6),(\mathrm{B} 5, \mathrm{R} 5),(\mathrm{B} 5, \mathrm{R} 6),(\mathrm{B} 6, \mathrm{R} 4),(\mathrm{B} 6, \mathrm{R} 5),(\mathrm{B} 6, \mathrm{R} 6)}

And \mathrm{B}={(\mathrm{B} 5, \mathrm{R} 1),(\mathrm{B} 5, \mathrm{R} 2),(\mathrm{B} 5, \mathrm{R} 3),(\mathrm{B} 5, \mathrm{R} 4),(\mathrm{B} 5, \mathrm{R} 5),(\mathrm{B} 5, \mathrm{R} 6)}

\Rightarrow A \cap B={(B 5, R 5),(B 5, R 6)}

So, \mathrm{P}(\mathrm{A})=\frac{6}{36}=\frac{1}{6}, \mathrm{P}(\mathrm{B})=\frac{6}{36}=\frac{1}{6}, \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{2}{36}=\frac{1}{18}

Now, we know that by definition of conditional probability,

\mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}

Now by substituting the values we get

\Rightarrow \mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{1 / 18}{1 / 6}=\frac{6}{18}=\frac{1}{3}

\Rightarrow \mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{1}{3}

(b) Let A be the event of ‘obtaining a sum 8^{\prime} and B be the event of ‘getting a number less than 4 on red die’. Then, A={(B 2, R 6),(B 3, R 5),(B 4, R 4),(B 5, R 3),(B 6, R 2)}

\mathrm{B}=\left{\begin{array}{l}(\mathrm{B} 1, \mathrm{R} 1)(\mathrm{B} 2, \mathrm{R} 1),(\mathrm{B} 3, \mathrm{R} 1),(\mathrm{B} 4, \mathrm{R} 1),(\mathrm{B} 5, \mathrm{R} 1),(\mathrm{B} 6, \mathrm{R} 1), \ (\mathrm{B} 1, \mathrm{R} 2),(\mathrm{B} 2, \mathrm{R} 2),(\mathrm{B} 3, \mathrm{R} 2),(\mathrm{B} 4, \mathrm{R} 2),(\mathrm{B} 5, \mathrm{R} 2),(\mathrm{B} 6, \mathrm{R} 2), \ (\mathrm{B} 1, \mathrm{R} 3),(\mathrm{B} 2, \mathrm{R} 3),(\mathrm{B} 3, \mathrm{R} 3),(\mathrm{B} 4, \mathrm{R} 3),(\mathrm{B} 5, \mathrm{R} 3),(\mathrm{B} 6, \mathrm{R} 3)\end{array}\right}

We have sample space of common event,

\Rightarrow A \cap B={(B 5, R 3),(B 6, R 2)}

The probability of the event,
P(A)=\frac{5}{36}, P(B)=\frac{18}{36}=\frac{1}{2}, P(A \cap B)=\frac{2}{36}=\frac{1}{18}

Now, we know that

By relation of conditional probability,

\mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}

Now by substituting the values we get

\Rightarrow \mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{1 / 18}{1 / 2}=\frac{2}{18}=\frac{1}{9}

\Rightarrow \mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{1}{9}