Three persons A, B,and C are to speak at a fu | vedclass

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Question

(B) The total number of ways in which $3$ persons $(A, B, C)$ can speak is $3! = 6$. These arrangements are: $(A, B, C), (A, C, B), (B, A, C), (B, C,

Vedclass · Accepted Answer

$1/6$

Vedclass · Answer

(B) The total number of ways in which $3$ persons $(A, B, C)$ can speak is $3! = 6$. These arrangements are: $(A, B, C), (A, C, B), (B, A, C), (B, C, A), (C, A, B), (C, B, A)$. We are looking for the condition where $A$ speaks before $B$ and $B$ speaks before $C$,which corresponds to the specific order $(A, B, C)$. There is only $1$ favorable outcome out of $6$ total possible outcomes. Therefore,the required probability is $\frac{1}{6}$.

Three persons $A, B$,and $C$ are to speak at a function. If they speak in a random order,what is the probability that $A$ speaks before $B$ and $B$ speaks before $C$?

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