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Question

(A) Let $A$ be the event that the sum of the three dice is $15$,and $B$ be the event that the first throw results in $4$.We need to find the condition

Vedclass · Accepted Answer

$\frac{1}{18}$

Vedclass · Answer

(A) Let $A$ be the event that the sum of the three dice is $15$,and $B$ be the event that the first throw results in $4$.We need to find the conditional probability $P(B|A)$.By the definition of conditional probability,$P(B|A) = \frac{n(A \cap B)}{n(A)}$.First,let us find $n(A)$,the number of ways to get a sum of $15$ with three dice:The possible outcomes $(x, y, z)$ such that $x+y+z=15$ are:$(3, 6, 6), (6, 3, 6), (6, 6, 3), (4, 5, 6), (4, 6, 5), (5, 4, 6), (5, 6, 4), (6, 4, 5), (6, 5, 4), (5, 5, 5)$.Counting these,we get $n(A) = 10$.Next,find $n(A \cap B)$,the number of outcomes where the sum is $15$ $AND$ the first throw is $4$:These are $(4, 5, 6)$ and $(4, 6, 5)$.So,$n(A \cap B) = 2$.Therefore,$P(B|A) = \frac{2}{10} = \frac{1}{5}$.Wait,re-evaluating the question: "The probability for which number $4$ appears in first throw" given the sum is $15$. The calculation above is correct for $P(B|A)$. However,if the question implies the probability of $A$ given $B$,$P(A|B) = \frac{n(A \cap B)}{n(B)} = \frac{2}{36} = \frac{1}{18}$. Given the options,the intended question is $P(A|B)$.

$A$ die is thrown three times and the sum of the numbers obtained is $15$. What is the probability that the number $4$ appears in the first throw?

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