Given that the two numbers appearing on throwing two dice are different, the probability of an event 'the sum of numbers on the dice is $4$ ' is:

\frac{1}{2}

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\frac{1}{15}

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\frac{1}{3}

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\frac{1}{5}

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Solution

The correct option is B $\frac{1}{15}$
The sample space of the given experiment is given below
$S = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6) (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6) (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) \end{matrix} ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭$
Let $E$ be the event that 'the sum of numbers on the dice is $4$ ' and $F$ be the event that 'the two numbers appearing on throwing the two dice are different.
$\Rightarrow E = {(1, 3), (2, 2), (3, 1)}$
$\Rightarrow F = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} (1, 2), (1, 3), (1, 4), (1, 5), (1, 6) (2, 1), (2, 3), (2, 4), (2, 5), (2, 6) (3, 1), (3, 2), (3, 4), (3, 5), (3, 6) (4, 1), (4, 2), (4, 3), (4, 5), (4, 6) (5, 1), (5, 2), (5, 3), (5, 4), (5, 6) (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), \end{matrix} ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭$
$\Rightarrow E \cap F = {(1, 3), (3, 1)}$
So, $P (E) = \frac{3}{36} = \frac{1}{12}, P (F) = \frac{30}{36} = \frac{5}{6}, P (E \cap F) = \frac{2}{36} = \frac{1}{18}$
Now, we know that, by definition of conditional probability,
$P (E / F) = \frac{P (E \cap F)}{P (F)}$
$\Rightarrow P (E / F) = \frac{1 / 18}{5 / 6} = \frac{1}{15}$
$\Rightarrow P (E / F) = \frac{1}{15}$

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Given that the two numbers appearing on throwing two dice are different. Find the probability of the event the sum of numbers on the dice is 4.

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