A die is rolled three times. The probability of getting their sum equal to a prime number of the form \(4…
A die is rolled three times. The probability of getting their sum equal to a prime number of the form \(4 n+1\) is
\(\frac{1}{6}\)
\(\frac{7}{36}\)
\(\frac{5}{36}\)
\(\frac{11}{36}\)
Solution
We obtain following prime sum in three throws of dice;
3, 5, 7, 11, 13 and 17
And sum in from of \(4 m+1\) are,
5, 13 and 17
Following are favourable outcomes for above sums
\(\begin{aligned}
& (1,2,2),(2,1,2),(2,2,1) \\
& (3,1,1),(1,3,1),(1,1,3) \\
& (5,3,3),(3,5,3),(3,3,5) \\
& (6,6,1),(1,6,6),(6,1,6) \\
& (4,4,5),(5,4,4),(4,5,4) \\
& (2,6,5),(6,2,5),(5,2,6)
\end{aligned}\)
.etc.
Total number of out comes of obtaining sum a prime of from ' \(4 m+\mathrm{l}^{\prime}=\operatorname{sum} 5(6\) times \()+\operatorname{sum} 13\) (21 times) + sum 17 ( 3 times) \(=30\) cares.
Total number of possible outcomes
\(=6 \times 6 \times 6=216 \text { cares }\)
Hence, required probability \(=\frac{30}{216}=\frac{5}{36}\).