ISI2018-DCG-6

Question

A die is thrown thrice. If the first throw is a $4$ then the probability of getting $15$ as the sum of three throws is

• A. $\frac{1}{108}$
• B. $\frac{1}{6}$
• C. $\frac{1}{18}$
• D. none of these

Comments

Answer - (C)

First throw is $4$. So, total no. of outcomes will be $6$ choices for second die and $6$ choices for the third die. 

So, total no.of outcomes

$n(S) = 6 \cdot 6 = 36$

Now, to get $15$ as sum, given that first die has $4$ on its face, it means the sum of faces of dice in second and third throw should be $15-4=11$

There are exactly $2$ such outcomes. 

$( 5,6)$ and $(6,5)$

Hence, no. of favourable outcomes $n(A) = 2$

Hence, required probability = $\frac{n(A)}{n(S)} = \frac{2}{36} = \frac{1}{18}$ 

Answer 1

Answer: $\mathrm {\bf C}$

Explanation:

We know that the first throw showed $4$.

Thus the second and third die throws must sum to $11$ in order to get the total sum as $15$

With six-sided dice, there are only two possible combinations for two throws to sum =  $11$

$(5,6)$ or $(6, 5)$

There are $36$ possible combinations and $2$ are favorable.
 Hence, probability = $\frac{2}{36}=\frac{1}{18}$

$\therefore \mathbf C$ is the correct option.