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In a single throw of three dice, the probability of getting a sum at least 5 is
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The correct answer is:
$\frac{53}{54}$
Here $n(S)=6 \times 6 \times 6=216$
Sum less than $5 \equiv\{(1,1,1),(1,1,2),(1,2,1),(2,1,1)\}$
Here $\mathrm{P}($ sum less than 5$)=\frac{4}{216}=\frac{1}{54}$
$\therefore \mathrm{P}($ at least 5$)=\mathrm{P}(\geq 5)=1-\mathrm{P}( < 5)$
$=1-\frac{1}{54}=\frac{53}{54}$
Sum less than $5 \equiv\{(1,1,1),(1,1,2),(1,2,1),(2,1,1)\}$
Here $\mathrm{P}($ sum less than 5$)=\frac{4}{216}=\frac{1}{54}$
$\therefore \mathrm{P}($ at least 5$)=\mathrm{P}(\geq 5)=1-\mathrm{P}( < 5)$
$=1-\frac{1}{54}=\frac{53}{54}$
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