Two dice are tossed. The probability that the total score is a prime number is-
A. $$\frac{1}{6}$$
B. $$\frac{1}{2}$$
C. $$\frac{5}{12}$$
D. $$\frac{7}{9}$$
E. None of these
Answer: Option C
Solution (By Examveda Team)
Clearly, n (S) = (6 × 6) = 36Let E be the event that the sum is a prime number. Then, n (E) = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3,), (5, 2), (5, 6), (6, 1), (6, 5)}
∴ n (E) = 15
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{15}}{{36}} = \frac{5}{{12}}$$
This Question Belongs to Arithmetic Ability >> Probability
Clearly, n (S) = (6 × 6) = 36
Let E be the event that the sum is a prime number. Then, n (E) = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3,), (5, 2), (5, 6), (6, 1), (6, 5)}
∴ n (E) = 15
∴P(E)=n(E)n(S)=1536=512
n (S) = (6 × 6) = 36 why we count 6 not choose than other number?