Three dice are thrown simultaneously . The probability that 4 appears on two dice , given that 5 has appeared on one dice is .
The way I’ve gone about it is :
_ _ 5 , _ 5 _ , 5 _ _
A: event of 4 appearing twice
B: event of 5 appearing once
(In three throws )
Then , P(A/B) = P(A∩B)/P(B)
= $\frac{^3C_1(1/6)^3}{^3C_1 (1/6)}$
Where $^3C_1$ in the numerator is for selecting any 1 dice for 5 and the other two automatically get selected for 4 .
While $(1/6)^3$ Is probability of digits 5 ,4,4 appearing on the 3 dices .
This method is not giving me the desired answer ,as given in the question .
The ans given is $\frac{3}{91}$
Best Answer
Your numerator is fine. Your problem is in the denominator. The probability of rolling at least one $5$ is equal to one minus the probability of rolling no $5$'s. This is
\begin{align} P(\text{at least one $5$ in three rolls}) & = 1-P(\text{no $5$ in three rolls}) \\ & = 1-[P(\text{no $5$ in one roll})]^3 \\ & = 1-\left(\frac56\right)^3 \\ & = 1-\frac{125}{216} = \frac{91}{216} \end{align}