Three dice are thrown. Given that the sum of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $S$ be the set of outcomes where the sum of three dice is $8$. The possible combinations (unordered) are:$(1, 1, 6) \rightarrow \frac{3!}{2!}

Vedclass · Accepted Answer

$\frac{3}{7}$

Vedclass · Answer

(B) Let $S$ be the set of outcomes where the sum of three dice is $8$. The possible combinations (unordered) are:$(1, 1, 6) \rightarrow \frac{3!}{2!} = 3$ permutations$(1, 2, 5) \rightarrow 3! = 6$ permutations$(1, 3, 4) \rightarrow 3! = 6$ permutations$(2, 2, 4) \rightarrow \frac{3!}{2!} = 3$ permutations$(2, 3, 3) \rightarrow \frac{3!}{2!} = 3$ permutationsTotal number of outcomes $n(S) = 3 + 6 + 6 + 3 + 3 = 21$.Let $E$ be the event that at least one die shows a $4$. The favorable outcomes are from $(1, 3, 4)$ and $(2, 2, 4)$.Number of favorable outcomes $n(E) = 6 + 3 = 9$.The probability $P(E) = \frac{n(E)}{n(S)} = \frac{9}{21} = \frac{3}{7}$.

Three dice are thrown. Given that the sum of the numbers on them is $8$,what is the probability that at least one of them shows a $4$?

Similar Questions

If $P(A)=0.8, P(B)=0.5$ and $P(B | A)=0.4,$ find $P(A \cup B).$

If $A$ and $B$ are two events such that $P(A) = 0.7$,$P(B) = 0.4$,and $P(A \cap \overline{B}) = 0.5$,where $\overline{B}$ denotes the complement of $B$,then $P(B \mid (A \cup \overline{B}))$ is equal to:

$A$ fair die is tossed until a six is obtained. Let $X$ be the number of required tosses,then the conditional probability $P(X \geq 5 \mid X > 2)$ is:

If two events $A$ and $B$ are such that $P(A^c) = 0.3$,$P(B) = 0.4$ and $P(A \cap B^c) = 0.5$,then $P(B | A \cup B^c)$ is equal to

If a die is rolled twice and the sum of the numbers appearing on them is observed to be $6$, then the probability that the number $1$ appears at least once on them is

Vedclass Test Series

Exam Paper Generator

Online Exam Module