Question

Two dice are thrown simultaneously. Find:

(a) P(an odd number as a sum) (b) P(sum as a prime number) (c) P(a doublet of odd numbers)

(d) P(a total of atleast 9)

(e) P( a multiple of 2 on one die and a multiple of 3 on other die)

(f) P(a doublet)

(g) P(a multiple of 2 as sum) (h) P(getting the sum 9) (i) P(getting a sum

greater than 12) (j) P( a prime number on each die) (k) P( a multiple of 5 as a sum)​

Answer 1

Answer:

P(an odd number as a sum) = 18/36

P(sum as a prime number) =15/36

P(a doublet of odd numbers) = 3/36

P(a total of at least 9) = 10/36

P( a multiple of 2 on one die and a multiple of 3 on other die) = 6/36

P(a doublet) = 6/36

P(a doublet) =18/36

P(getting the sum 9) = 4/36

P(getting a sum greater than 12) = 0/36

P( a prime number on each die) = 6/36

P( a multiple of 5 as a sum)​ = 7/36

Step-by-step explanation:

Total output = 36

a. (1,2) (1,4) (1,6) (2,1) (2,3) (2,5) (3,2) (3,4) (3,6) (4,1) (4,3) (4,5) (5,2) (5,4) (5,6)            

   (6,1) (6,3) (6,5)

   P(an odd number as a sum) = 18/36

b. (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)

   P(sum as a prime number) =15/36

c. (1,1) (3,3) (5,5)

    P(a doublet of odd numbers) = 3/36

d. (3,6) (4,5) (4,6) (5,4) (5,5) (5,6) (6,3) (6,4) (6,5) (6,6)

   P(a total of at least 9) = 10/36

e. (2,3) (2,6) (4,3) (4,6) (6,3) (6,6)

   P( a multiple of 2 on one die and a multiple of 3 on other die) = 6/36

f. (1,1) (2,2) (3,3) (4,4) (5,5) (6,6)

   P(a doublet) = 6/36

g. (1,1) (1,3) (2,2) (3,1) (1,5) (2,4) (3,3) (4,2) (5,1) (2,6) (3,5) (4,4) (5,3) (6,2) (4,6)

   (5,5) (6,4) (6,6)

   P(a doublet) =18/36

h. (3,6) (4,5) (5,4) (6,3)

   P(getting the sum 9) = 4/36

i. nil [ since max sum = 12 for (6,6) ]

   P(getting a sum greater than 12) = 0/36

j. (2,3) (2,5) (3,2) (3,5) (5,2) (5,5)

   P( a prime number on each die) = 6/36

k. (1,4) (2,3) (3,2) (4,1) (4,6) (5,5) (6,4)

   P( a multiple of 5 as a sum)​ = 7/36