Probability Of Rolling Two Dice

Probability for Rolling Ii Die

Probability for rolling two die with the six sided dots such as one, 2, 3, 4, 5 and 6 dots in each dice.

$Probability of Rolling Two Dice$

When two dice are thrown simultaneously, thus number of outcome can exist 6ⁱⁱ = 36 considering each die has 1 to 6 number on its faces. Then the possible outcomes are shown in the below tabular array.

Probability – Sample space for two dice (outcomes):

$Probability for Rolling Two Dice$

Note:

(i) The outcomes (1, 1), (2, two), (3, 3), (4, four), (5, 5) and (6, 6) are called doublets.

(ii) The pair (i, two) and (ii, 1) are unlike outcomes.

Worked-out issues involving probability for rolling two dice:

i. Ii dice are rolled. Let A, B, C exist the events of getting a sum of two, a sum of iii and a sum of iv respectively. Then, show that

(i) A is a unproblematic result

(two) B and C are compound events

(iii) A and B are mutually exclusive

Solution:

Conspicuously, nosotros have
A = {(1, one)}, B = {(one, 2), (two, i)} and C = {(ane, 3), (3, 1), (2, ii)}.

(i) Since A consists of a single sample indicate, it is a simple result.

(ii) Since both B and C contain more than one sample point, each one of them is a compound event.

(iii) Since A ∩ B = ∅, A and B are mutually exclusive.

ii. Two dice are rolled. A is the upshot that the sum of the numbers shown on the two die is 5, and B is the effect that at to the lowest degree one of the dice shows upwardly a 3.
Are the two events (i) mutually sectional, (2) exhaustive? Give arguments in support of your answer.

Solution:

When two dice are rolled, we have n(S) = (6 × 6) = 36.

At present, A = {(i, 4), (2, three), (4, one), (3, two)}, and

B = {(iii, i), (iii, 2), (3, three), (3, 4), (3, 5), (3, six), (i,three), (2, 3), (4, iii), (v, three), (6, 3)}

(i) A ∩ B = {(2, 3), (3, 2)} ≠ ∅.

Hence, A and B are non mutually exclusive.

(two) Also, A ∪ B ≠ Southward.

Therefore, A and B are not exhaustive events.

More examples related to the questions on the probabilities for throwing ii dice.

3. Ii dice are thrown simultaneously. Detect the probability of:

(i) getting half dozen as a production

(ii) getting sum ≤ iii

(iii) getting sum ≤ 10

(iv) getting a doublet

(5) getting a sum of viii

(vi) getting sum divisible by 5

(seven) getting sum of atleast 11

(eight) getting a multiple of 3 as the sum

(ix) getting a total of atleast 10

(10) getting an even number as the sum

(11) getting a prime as the sum

(xii) getting a doublet of even numbers

(13) getting a multiple of 2 on 1 die and a multiple of three on the other die

Solution:

Two dissimilar dice are thrown simultaneously beingness number 1, 2, 3, four, 5 and half-dozen on their faces. We know that in a single thrown of two dissimilar dice, the total number of possible outcomes is (6 × 6) = 36.

(i) getting six equally a product:

Let Due east₁ = upshot of getting half dozen equally a product. The number whose product is six will be E_ane = [(i, 6), (2, iii), (3, 2), (6, 1)] = four

Therefore, probability of getting 'six as a product'

Number of favorable outcomes
P(E₁) = Total number of possible upshot

= iv/36
= ane/9

(ii) getting sum ≤ 3:

Permit E_two = event of getting sum ≤ 3. The number whose sum ≤ 3 will exist E₂ = [(1, 1), (1, two), (2, 1)] = iii

Therefore, probability of getting 'sum ≤ three'

Number of favorable outcomes
P(Due east₂) = Total number of possible upshot

= iii/36
= 1/12

(iii) getting sum ≤ 10:

Let E₃ = event of getting sum ≤ x. The number whose sum ≤ ten will exist Due east₃ =

[(1, i), (1, two), (1, 3), (i, 4), (1, five), (one, six),

(2, ane), (2, 2), (2, 3), (ii, 4), (2, 5), (2, 6),

(3, one), (3, 2), (3, 3), (three, 4), (3, v), (3, vi),

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

(five, 1), (5, 2), (5, 3), (5, four), (5, 5),

(6, 1), (6, 2), (half-dozen, 3), (six, 4)] = 33

Therefore, probability of getting 'sum ≤ ten'

Number of favorable outcomes
P(Due east_three) = Total number of possible result

= 33/36
= 11/12

(iv) getting a doublet: Allow E_four = event of getting a doublet. The number which doublet volition exist E₄ = [(1, 1), (ii, 2), (3, 3), (4, 4), (5, v), (6, 6)] = half-dozen

Therefore, probability of getting 'a doublet'

Number of favorable outcomes
P(E₄) = Total number of possible event

= half-dozen/36
= ane/6

(five) getting a sum of viii:

Permit E₅ = outcome of getting a sum of eight. The number which is a sum of 8 will exist E_v = [(ii, six), (3, 5), (4, 4), (5, 3), (six, 2)] = 5

Therefore, probability of getting 'a sum of 8'

Number of favorable outcomes
P(East₅) = Total number of possible outcome

= v/36

(half-dozen) getting sum divisible past 5:

Let East_half-dozen = issue of getting sum divisible past 5. The number whose sum divisible by 5 will be Due east₆ = [(i, 4), (2, iii), (3, 2), (4, 1), (4, vi), (5, v), (vi, 4)] = vii

Therefore, probability of getting 'sum divisible past v'

Number of favorable outcomes
P(E₆) = Total number of possible event

= 7/36

(vii) getting sum of atleast eleven:

Permit E₇ = upshot of getting sum of atleast xi. The events of the sum of atleast eleven will be Due east₇ = [(five, half-dozen), (half-dozen, v), (half dozen, six)] = three

Therefore, probability of getting 'sum of atleast 11'

Number of favorable outcomes
P(East₇) = Total number of possible issue

= iii/36
= one/12

(eight) getting a multiple of three as the sum:

Let E₈ = event of getting a multiple of 3 equally the sum. The events of a multiple of 3 every bit the sum will be E₈ = [(1, 2), (1, five), (ii, one), (2, 4), (three, 3), (three, six), (4, 2), (4, 5), (five, 1), (v, 4), (6, iii) (6, 6)] = 12

Therefore, probability of getting 'a multiple of three as the sum'

Number of favorable outcomes
P(E₈) = Total number of possible outcome

= 12/36
= 1/three

(ix) getting a total of atleast x:

Allow E₉ = event of getting a total of atleast x. The events of a full of atleast ten volition be E_ix = [(4, 6), (5, five), (five, 6), (vi, four), (6, 5), (six, 6)] = half dozen

Therefore, probability of getting 'a total of atleast 10'

Number of favorable outcomes
P(E₉) = Total number of possible outcome

= six/36
= 1/half-dozen

(10) getting an even number equally the sum:

Let East₁₀ = event of getting an even number as the sum. The events of an even number as the sum will be E_ten = [(1, 1), (1, iii), (1, 5), (ii, 2), (ii, 4), (two, half dozen), (3, 3), (iii, 1), (three, 5), (4, 4), (4, 2), (iv, 6), (v, ane), (v, 3), (5, v), (6, ii), (half-dozen, 4), (6, half dozen)] = 18

Therefore, probability of getting 'an even number equally the sum

Number of favorable outcomes
P(Eastward₁₀) = Full number of possible result

= 18/36
= 1/ii

(xi) getting a prime number as the sum:

Let Due east_eleven = event of getting a prime number as the sum. The events of a prime number every bit the sum will be E₁₁ = [(one, 1), (1, ii), (ane, 4), (1, half dozen), (2, 1), (2, 3), (2, five), (iii, two), (3, four), (4, 1), (four, iii), (five, 2), (5, 6), (six, 1), (6, 5)] = 15

Therefore, probability of getting 'a prime number as the sum'

Number of favorable outcomes
P(E_xi) = Full number of possible effect

= fifteen/36
= 5/12

(xii) getting a doublet of even numbers:

Allow Eastward₁₂ = result of getting a doublet of even numbers. The events of a doublet of even numbers volition exist E₁₂ = [(2, 2), (iv, 4), (6, 6)] = 3

Therefore, probability of getting 'a doublet of even numbers'

Number of favorable outcomes
P(E₁₂) = Total number of possible effect

= 3/36
= i/12

(13) getting a multiple of 2 on one die and a multiple of three on the other dice:

Let Eastward₁₃ = event of getting a multiple of two on one die and a multiple of three on the other die. The events of a multiple of 2 on one die and a multiple of 3 on the other die will exist E₁₃ = [(2, 3), (two, half dozen), (3, 2), (three, 4), (three, half dozen), (4, 3), (iv, half dozen), (6, 2), (half-dozen, 3), (half-dozen, 4), (vi, six)] = 11

Therefore, probability of getting 'a multiple of 2 on one dice and a multiple of 3 on the other die'

Number of favorable outcomes
P(E₁₃) = Total number of possible outcome

= 11/36

iv. Ii dice are thrown. Find (i) the odds in favour of getting the sum v, and (ii) the odds against getting the sum 6.

Solution:

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

Let Southward be the sample space. Then, north(Due south) = 36.

(i) the odds in favour of getting the sum 5:

Permit Due east₁ be the event of getting the sum 5. Then,
E₁ = {(one, 4), (2, 3), (three, ii), (four, 1)}
⇒ P(East₁) = 4
Therefore, P(E₁) = north(E_ane)/n(S) = four/36 = 1/9
⇒ odds in favour of E_one = P(E_i)/[1 – P(Eastward_i)] = (1/9)/(one – 1/9) = 1/8.

(two) the odds against getting the sum half dozen:

Let Due east₂ be the outcome of getting the sum half dozen. Then,
East₂ = {(ane, v), (two, iv), (three, three), (four, two), (5, 1)}
⇒ P(East₂) = 5
Therefore, P(E_ii) = n(E_ii)/due north(South) = 5/36
⇒ odds against E₂ = [i – P(E₂)]/P(Eastward_two) = (one – 5/36)/(5/36) = 31/5.

5. Two dice, one blue and ane orange, are rolled simultaneously. Find the probability of getting

(i) equal numbers on both

(2) ii numbers actualization on them whose sum is 9.

Solution:

The possible outcomes are

(ane, 1), (1, 2), (1, three), (1, 4), (i, 5), (1, vi),

(2, 1), (ii, two), (2, 3), (2, 4), (2, five), (two, 6),

(3, 1), (3, ii), (3, 3), (3, four), (3, 5), (3, half-dozen),

(4, i), (4, 2), (iv, 3), (4, four), (4, five), (iv, half-dozen)

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

(6, i), (vi, 2), (half-dozen, 3), (6, 4), (6, five), (half-dozen, 6)

$Sample Space of Rolling a Pair of Dice$

Therefore, total number of possible outcomes = 36.

(i) Number of favourable outcomes for the event E

= number of outcomes having equal numbers on both dice

= 6 [namely, (ane, 1), (ii, 2), (3, iii), (four, 4), (five, 5), (6, vi)].

So, by definition, P(Due east) = $\frac{vi}{36}$

= $\frac{i}{vi}$

(2) Number of favourable outcomes for the event F

= Number of outcomes in which 2 numbers appearing on them accept the sum 9

= four [namely, (3, half dozen), (4, 5), (5, 4), (3, 6)].

Thus, by definition, P(F) = $\frac{4}{36}$

= $\frac{ane}{nine}$.

These examples volition help us to solve different types of issues based on probability for rolling ii die.

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Probability of Tossing Three Coins

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Conditional Probability

Theoretical Probability

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Probability for Rolling Two Dice

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