Question

Counters numbered $1,2,3$ are placed in a bag and one is drawn at random and replaced. The operation is being repeated three times. The probabolity of obtaining a total of $6$ is ?

• A. $\frac{7}{9}$
• B. $\frac{6}{27}$
• C. $\frac{7}{27}$
• D. $\frac{5}{27}$

Answer 1

Answer: (C)

$\frac{7}{27}$

$Togetasumof6,thethreenumberscouldbeeither(1,2,3)or(2,2,2)$

$For(1,2,3),thetotalnumberofdifferentarrangements{}_3^{3}p=\frac{3}{0}=6$

$Totalnumberofpossibilitiesthatcouldoccurin3operations=3\ast 3\ast 3=27$

$Probabilityofobtaininga1inoneoperation=Probabilityofobtaininga2inoneoperation=Probabilityofobtaininga3inoneoperation=\frac{1}{3}$

$Probabilityofobtaining1and2and3inanyorder6\times \frac{1}{3}.\frac{1}{3}.\frac{1}{3}=\frac{6}{27}$

$Probabilityofobtaining2ineachofthe3operations\frac{1}{3}\frac{1}{3}.\frac{1}{3}=\frac{1}{27}$

$Chanceorprobabilityofobtainingatotalof6=\frac{6}{27}+\frac{1}{27}=\frac{7}{27}$