Question

A die is rolled three times. The probability of getting their sum equal to a prime number of the form $4n+1$ is__________

• A. $\frac{1}{6}$
• B. $\frac{7}{36}$
• C. $\frac{5}{36}$
• D. $\frac{11}{36}$

Answer 1

Answer: (C)

We obtain following prime sum in three throws of dice;
$3,5,7,11,13$ and $17$
And sum in from of $4m+1$ are, $5,13$ and $17$
Following are favourable outcomes for above sums
$(1,2,2),(2,1,2),(2,2,1)$
$(3,1,1),(1,3,1),(1,1,3)$
$(5,3,3),(3,5,3),(3,3,5)$
$(6,6,1),(1,6,6),(6,1,6)$
$(4,4,5),(5,4,4),(4,5,4)$
$(2,6,5),(6,2,5),(5,2,6)$
$\dots \dots \dots .$ etc.
Total number of out comes of obtaining sum a prime of from $4m+1=$ sum $5(6$ times $)+$ sum $13(21$ times $)+$ sum $17(3$ times $)=30$ cares.
Total number of possible outcomes $=6\times 6\times 6=216$ cares
Hence, required probability $=\frac{30}{216}=\frac{5}{36}$