Dice Probability Formulas for Competitive Exams

Formulas of Dice Probability Questions

Example.3. Two dice are thrown simultaneously. Find the probability of:

(a)  Getting sum of at least 11

Solution:

Two different dice are thrown simultaneously being number 1, 2, 3, 4, 5 and 6 on their faces. We know that in a single thrown of two different dice, the total number of possible outcomes is (6 × 6) = 36

getting sum of atleast 11:

Let E₇ = event of getting sum of atleast 11. The events of the sum of atleast 11 will be E₇ = [(5, 6), (6, 5), (6, 6)] = 3

Therefore, probability of getting ‘sum of atleast 11’

$$P(E) ={Number \ of \ favorable \ outcomes\over Total \ number \ of \ possible \ outcome }$$
       = 3/36
       = 1/12

(b)  getting an even number as the sum

Solution:

Two different dice are thrown simultaneously being number 1, 2, 3, 4, 5 and 6 on their faces. We know that in a single thrown of two different dice, the total number of possible outcomes is (6 × 6) = 36

getting an even number as the sum:

Let E₁₀ = event of getting an even number as the sum. The events of an even number as the sum will be E₁₀ = [(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 3), (3, 1), (3, 5), (4, 4), (4, 2), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)] = 18

Therefore, probability of getting ‘an even number as the sum

$$P(E) ={Number \ of \ favorable \ outcomes\over Total \ number \ of \ possible \ outcome }$$
 
       = 18/36
       = 1/2

Example.4. Two dice are thrown.

Find (a) the odds in favour of getting the sum 5

Solution:

We know that in a single thrown of two die, the total number of possible outcomes is (6 × 6) = 36.

Let S be the sample space. Then, n(S) = 36.

Let E₁ be the event of getting the sum 5. Then,
 E₁ = {(1, 4), (2, 3), (3, 2), (4, 1)}
 ⇒ P(E₁) = 4
 Therefore, P(E₁) = n(E₁)/n(S) = 4/36 = 1/9
 ⇒ odds in favour of E₁ = P(E₁)/[1 – P(E₁)] = (1/9)/(1 – 1/9) = 1/8.
 

Find (b) the odds against getting the sum 6.

Solution:

We know that in a single thrown of two die, the total number of possible outcomes is (6 × 6) = 36.

Let S be the sample space. Then, n(S) = 36.

Let E₂ be the event of getting the sum 6. Then,
 E₂ = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)} 
 ⇒ P(E₂) = 5
 Therefore, P(E₂) = n(E₂)/n(S) = 5/36 
 ⇒ odds against E₂ = [1 – P(E₂)]/P(E₂) = (1 – 5/36)/(5/36) = 31/5.

Probability for Rolling Three Dice

Probability for rolling three dice with the six sided dots such as 1, 2, 3, 4, 5 and 6 dots in each (three) dies.

When three dice are thrown simultaneously/randomly, thus number of event can be 6³ = (6 × 6 × 6) = 216 because each die has 1 to 6 number on its faces.

Example: Three dice are thrown together. Find the probability of:

(a) Getting a total of 6.

Solution:

Three different dice are thrown at the same time.

Therefore, total number of possible outcomes will be 6³ = (6 × 6 × 6) = 216.

getting a total of 6:

Number of events of getting a total of 6 = 10

i.e. (1, 1, 4), (1, 4, 1), (4, 1, 1), (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1) and (2, 2, 2).

Therefore, probability of getting a total of 6

$$P(E) ={Number \ of \ favorable \ outcomes\over Total \ number \ of \ possible \ outcome }$$
  = 10/216

= 5/108.

(b) Getting a total of 5.

Solution:

Three different dice are thrown at the same time.

Therefore, total number of possible outcomes will be 6³ = (6 × 6 × 6) = 216.

getting a total of 5:

Number of events of getting a total of 5 = 6

i.e. (1, 1, 3), (1, 3, 1), (3, 1, 1), (2, 2, 1), (2, 1, 2) and (1, 2, 2)

Therefore, probability of getting a total of 5

$$P(E) ={Number \ of \ favorable \ outcomes\over Total \ number \ of \ possible \ outcome }$$
  = 6/216
  = 1/36

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