A black and a red die are rolled together. Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

Solution

Here $S = \{\begin{matrix} (1, 1) & (1, 2) & (1, 3) & (1, 4) & (1, 5) & (1, 6) \\ (2, 1) & (2, 2) & (2, 3) & (2, 4) & (2, 5) & (2, 6) \\ (3, 1) & (3, 2) & (3, 3) & (3, 4) & (3, 5) & (3, 6) \\ (4, 1) & (4, 2) & (4, 3) & (4, 4) & (4, 5) & (4, 6) \\ (5, 1) & (5, 2) & (5, 3) & (5, 4) & (5, 5) & (5, 6) \\ (6, 1) & (6, 2) & (6, 3) & (6, 4) & (6, 5) & (6, 6) \end{matrix}\}$

A number of elements (outcomes) of the above sample space is 6 × 6 = 36.
Let E: a total of 8

$∴ E : \{(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)\} F : red die resulted in a number less than 4 ∴ F = \{(1, 1), (1, 2), . . . ., (1, 6), (2, 1), (2, 2), . . . ., (2, 6), (3, 1), (3, 2), . . . . (3, 6)\} ∴ E \cap F = \{(2, 6), (3, 5)\} ∴ P (F) = \frac{18}{36}, P (E \cap F) = \frac{2}{36}$ $∴ required probability = P (E F) = \frac{P (E \cap F)}{P (F)} = \frac{\frac{2}{36}}{\frac{18}{36}} = \frac{2}{18} = \frac{1}{9}$

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Probability

A black and a red die are rolled together. Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

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