PROBABILITY PROBLEMS ON DICE

Problem 1:

Two dice are thrown together. Find the probability that the product of the numbers on the top of the dice is

(i) 6      (ii) 12       (iii) 7

Solution :

i)

Total number of outcomes = 6 × 6 = 36

Favorable outcomes = {(1, 6), (2, 3), (3, 2), (6, 1)}

P (getting a 6) = 4/36

= 1/9

ii)

Favorable are of getting the product on the top of the dice

= {(2, 6) (6, 2) (4, 3) (3, 4)}

Total favorable outcomes = 4

P (getting the product of the no on the top of dice 12)

= 4/36

= 1/9

iii)

Favorable cases of getting the product of no on the top of dice 7 = 0

P (getting the product of the no on the top of dice 7) = 0/36

= 0

Problem 2 :

Two dice are thrown at the same time and the product of numbers appearing on them is noted. Find the probability that the product is less than 9.

Solution :

Number of total outcomes = 36

Favorable outcomes

=  {(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (3, 1) (3, 2)(4, 1) (4, 2) (5, 1) (6, 1) }

Number of possible ways = 16

P (the product is less than 9) = 16/36

= 4/9

Problem 3 :    

Two dice are numbered 1, 2, 3, 4, 5, 6 and 1, 1, 2, 2, 3, 3, respectively. They are thrown and the sum of the numbers on them is noted. Find the probability of getting each sum from 2 to 9 separately.

Solution :

Number of total outcomes = 36

Event of getting sum 2 = {(1, 1) (1, 1)}

P (getting sum 2) = 2/36

= 1/18

Event of getting sum 3 = {(1, 2) (1, 2) (2, 1) (2, 1)}

P (getting sum 3) = 4/36

= 1/9

Event of getting sum 4 = {(2, 2) (3, 1) (1, 3)}

P (getting sum 4) = 3/36

= 1/12

Event of getting sum 5 = { (1, 4) (2, 3) (3, 2) (4, 1) }

P (getting sum 5) = 4/36

= 1/9

Event of getting sum 6 = {(1, 5) (2, 4) (3, 3) (4, 2) (5, 1)} 

P (getting sum 6) = 5/36

Event of getting sum 7 = {(1, 6) (2, 5)(3, 4) (4, 3)(5, 2) (6, 1)}

P (getting sum 7) = 6/36

= 1/6

Event of getting sum 8 = {(2, 6) (3, 5) (4, 4)(5, 3) (6, 2)}

P (getting sum 8) = 5/36

Event of getting sum 9 = {(3, 6))5, 4) (4, 5) (6, 3)}

P (getting sum 9) = 4/36

= 1/9

Problem 4 :

Two dice are thrown at the same time. Determine the probability that the difference of the numbers on the two dice is 2.

Solution :

When two dice are thrown, we get 36 result.

(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)

Difference is 2 result is

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

Probability P(A) = 8/36

= 2/9

Problem 5 :

Two dice are thrown at the same time. Find the probability of getting (i) same number on both dice (ii) different numbers on both dice.     

Solution :

(i)           Same number on both dice

A {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

P (getting same number) = 6/36

= 1/6

(ii)  Different numbers on both dice

A {(1, 1), (1, 2), (1, 3), (1, 4)…... (6, 2)}

P (getting different number) = 30/36

= 5/6

Problem 6 :

Two dice are thrown simultaneously. What is the probability that the sum of the numbers appearing on the dice is

(i) 7?    (ii) a prime number?   (iii) 1?

Solution :

(i)           Sum of numbers appearing on the dice is 7.

So, the possible outcomes = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

Number of possible outcomes = 6

P (Sum of numbers appearing on the dice is 7) = 6/36

= 1/6

(ii) Sum of the numbers appearing on the dice is a prime number 2, 3, 5, 7, 11.

So, the possible outcomes = {(1, 1), (1, 2), (2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (5, 6), (6, 5)}

Number of possible outcomes = 15

P (a prime number) = 15/36

= 5/12

(iii) Sum of numbers appearing on the dice is 1.

It is not possible, so its probability is zero.

Problem 7:

Two dice are thrown simultaneously. Find the probability of getting

a. An even number on first dice

b. An odd number on first dice

c. An even number as the sum

d. A multiple of 5 as the sum

e. A multiple of 7 as the sum

f. A multiple of 3 as the sum

g. A sum more then 7

h. A sum greater than 9

Solution :

a.

An even number on first dice

Favorable outcomes to even number on first dice

{(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Number of possible outcomes = 18

P (an even number on first dice) = 18/36

= 1/2

b.

An odd number on first dice

Favorable outcomes to odd number on first dice

{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}

Number of possible outcomes = 18

P (an odd number on first dice) = 18/36

= 1/2

c.

An even number as the sum

Favorable outcomes to an even number as the sum on dice

{(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)}

Number of possible outcomes = 18

P (an even number as the sum) = 18/36

= 1/2

d.

A multiple of 5 as the sum

Favorable outcomes to the multiple of 5 as the sum on dice

{(1, 4), (2, 3), (3, 2), (4, 1), (4, 6), (5, 5), (6, 4)}

Number of possible outcomes = 7

P (A multiple of 5 as the sum) = 7/36

e.

A multiple of 7 as the sum

Favorable outcomes to the multiple of 7 as the sum on dice

{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

Number of possible outcomes = 6

P (A multiple of 5 as the sum) = 6/36

= 1/6

f.

A multiple of 3 as the sum

Favorable outcomes to the multiple of 3 as the sum on dice

{(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 possible outcomes = 12

P (a multiple of 3 as the sum) = 12/36

= 1/3

g.

A sum more then 7

Favorable outcomes = {(2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Number of possible outcomes = 15

P (getting a number more than 7) = 15/36

= 5/12

h.

A sum greater than 9

Favorable outcomes = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}

Number of possible outcomes = 6

P (getting a number more than 9) = 6/36

= 1/6