Question

Two dice are thrown simultaneously. Find the probability of getting:
(i)The sum as a prime number.
(ii) A total of at least 10.
(iii) A doublet of even number.
(iv) A multiple of 2 on on. dice and a multiple of 3 on the other dice.
(v) A multiple of 3 am the sum.

Answer 1

Sample space for total number of possible outcomes

$(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),$

$(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),$

$(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),$

$(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),$

$(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),$

$(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)$

Total number of outcomes $=36$

(i)

Favorable outcomes for sum as prime are

$(1,1),(1,2),(1,4),(1,6),(2,3),(2,3),(2,5),(3,2),(3,4),(4,1),(4,3),(5,2),(5,6),(6,1),(6,5)$

Number of favorable outcomes $=15$

Hence, the probability of getting the sum as a prime number. = $\frac{15}{36}=\frac{5}{12}$

(ii)

Favorable outcomes for total of atleast 10 are

$(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)$

Number of favorable outcomes $=6$

Hence, the probability of getting a total of atleast 10 = $\frac{6}{36}=\frac{1}{6}$

(iii)

Favorable outcomes for a doublet of even number are

$(2,2),(4,4),(6,6)$

Number of favorable outcomes $=3$

Hence, the probability of getting a doublet of even number = $\frac{3}{36}=\frac{1}{13}$

(iv)

Favorable outcomes for a multiple of 2 on one dice and a multiple of 3 on the other dice are

$(2,3),(2,6),(3,2),(3,4),(3,6),(4,3),(4,6),(6,2),(6,3),(6,4),(6,6)$

Number of favorable outcomes $=11$

Hence, the probability of getting a multiple of 2 on one dice and a multiple of 3 on the other dice = $\frac{11}{36}$

(v)

Favorable outcomes for getting a multiple of 3 as the sum

$(1,2),(1,5),(2,1),(2,4),(3,3),(3,6),(4,2),(4,5),(5,1),(5,4),(6,3)(6,6)$

Number of favorable outcomes $=12$

Hence, the probability of getting a multiple of $3$ as the sum = $\frac{12}{36}=\frac{1}{3}$