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A Family Has Five Kids. What Is the Probability That They Consist of 2 Boys and 3 Girls

Probability for Rolling Two Die

Probability for rolling two dice with the six sided dots such as 1, 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 be 62 = 36 because each die has 1 to 6 number on its faces. Then the possible outcomes are shown in the beneath tabular array.

Probability – Sample infinite for 2 dice (outcomes):

Probability for Rolling Two Dice

Note:

(i) The outcomes (1, ane), (2, two), (three, 3), (4, 4), (v, v) and (6, vi) are called doublets.

(ii) The pair (one, 2) and (two, 1) are dissimilar outcomes.

Worked-out bug involving probability for rolling two dice:

1. Ii dice are rolled. Let A, B, C be the events of getting a sum of 2, a sum of 3 and a sum of four respectively. Then, testify that

(i) A is a simple event

(ii) B and C are chemical compound events

(3) A and B are mutually exclusive

Solution:

Clearly, we have
A = {(1, 1)}, B = {(1, 2), (2, 1)} and C = {(ane, 3), (3, 1), (two, 2)}.

(i) Since A consists of a single sample point, information technology is a simple event.

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

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

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

Solution:

When two dice are rolled, we have northward(South) = (vi × half dozen) = 36.

Now, A = {(1, 4), (2, 3), (4, 1), (3, 2)}, and

B = {(3, 1), (iii, two), (three, iii), (3, 4), (3, 5), (three, 6), (1,3), (ii, 3), (4, 3), (v, iii), (6, 3)}

(i) A ∩ B = {(2, three), (3, two)} ≠ ∅.

Hence, A and B are not mutually sectional.

(two) Also, A ∪ B ≠ South.

Therefore, A and B are not exhaustive events.

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

iii. Two dice are thrown simultaneously. Find the probability of:

(i) getting six as a production

(ii) getting sum ≤ 3

(iii) getting sum ≤ x

(4) getting a doublet

(5) getting a sum of viii

(half-dozen) getting sum divisible past v

(7) getting sum of atleast 11

(viii) getting a multiple of 3 equally the sum

(9) getting a full of atleast ten

(10) getting an even number as the sum

(eleven) getting a prime number number as the sum

(xii) getting a doublet of even numbers

(thirteen) getting a multiple of 2 on one die and a multiple of 3 on the other die

Solution:

Ii unlike dice are thrown simultaneously being number one, 2, iii, 4, 5 and half dozen on their faces. We know that in a unmarried thrown of two different die, the total number of possible outcomes is (half dozen × 6) = 36.

(i) getting 6 as a product:

Let E1 = result of getting six as a product. The number whose product is six will be Eone = [(one, half-dozen), (2, 3), (three, 2), (6, 1)] = four

Therefore, probability of getting 'six every bit a product'

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

= 4/36
= one/9

(two) getting sum ≤ iii:

Permit E2 = event of getting sum ≤ 3. The number whose sum ≤ 3 will be E2 = [(ane, i), (1, 2), (2, ane)] = 3

Therefore, probability of getting 'sum ≤ 3'

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

= 3/36
= 1/12

(iii) getting sum ≤ ten:

Permit Due eastiii = outcome of getting sum ≤ ten. The number whose sum ≤ x will be Eastiii =

[(1, 1), (1, 2), (i, iii), (1, iv), (1, 5), (1, 6),

(2, i), (2, ii), (2, 3), (2, iv), (2, 5), (ii, 6),

(iii, ane), (three, two), (three, 3), (3, iv), (iii, 5), (three, 6),

(4, 1), (4, 2), (4, 3), (4, iv), (4, 5), (4, six)

(5, 1), (5, 2), (five, 3), (v, iv), (5, 5),

(vi, ane), (6, 2), (6, 3), (half dozen, four)] = 33

Therefore, probability of getting 'sum ≤ ten'

               Number of favorable outcomes
P(E3) = Full number of possible outcome

= 33/36
= 11/12

(iv) getting a doublet: Permit E4 = issue of getting a doublet. The number which doublet will be Eiv = [(1, 1), (2, 2), (3, 3), (four, four), (five, 5), (6, 6)] = vi

Therefore, probability of getting 'a doublet'

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

= 6/36
= i/vi

(v) getting a sum of 8:

Let Due eastfive = result of getting a sum of 8. The number which is a sum of 8 will be E5 = [(2, 6), (3, 5), (four, 4), (5, 3), (vi, ii)] = 5

Therefore, probability of getting 'a sum of 8'

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

= 5/36

(vi) getting sum divisible past 5:

Allow Ehalf-dozen = event of getting sum divisible past 5. The number whose sum divisible by 5 will be E6 = [(1, four), (2, 3), (3, two), (4, 1), (4, six), (5, 5), (six, four)] = seven

Therefore, probability of getting 'sum divisible past 5'

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

= 7/36

(vii) getting sum of atleast eleven:

Let E7 = result of getting sum of atleast 11. The events of the sum of atleast 11 will be E7 = [(five, 6), (half dozen, v), (6, half dozen)] = three

Therefore, probability of getting 'sum of atleast 11'

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

= three/36
= 1/12

(8) getting a multiple of 3 as the sum:

Let Eeight = outcome of getting a multiple of 3 as the sum. The events of a multiple of 3 as the sum will be E8 = [(ane, 2), (1, 5), (two, ane), (2, iv), (three, 3), (3, vi), (4, 2), (4, 5), (5, 1), (v, 4), (vi, iii) (six, six)] = 12

Therefore, probability of getting 'a multiple of 3 equally the sum'

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

= 12/36
= 1/3

(ix) getting a total of atleast 10:

Permit Eix = effect of getting a total of atleast x. The events of a total of atleast ten volition exist Due east9 = [(iv, 6), (5, five), (5, 6), (6, 4), (6, 5), (6, 6)] = 6

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

               Number of favorable outcomes
P(East9) = Total number of possible outcome

= vi/36
= 1/6

(x) getting an even number every bit the sum:

Permit Eastward10 = event of getting an fifty-fifty number as the sum. The events of an even number every bit the sum will be E10 = [(one, i), (i, iii), (one, five), (2, ii), (2, iv), (2, 6), (iii, iii), (three, 1), (three, 5), (4, 4), (4, 2), (4, six), (v, ane), (5, 3), (5, 5), (half dozen, 2), (6, 4), (6, half-dozen)] = 18

Therefore, probability of getting 'an fifty-fifty number as the sum

               Number of favorable outcomes
P(E10) = Full number of possible outcome

= eighteen/36
= 1/two

(xi) getting a prime number as the sum:

Let Exi = issue of getting a prime number number as the sum. The events of a prime number as the sum will be Exi = [(ane, 1), (ane, two), (1, 4), (one, six), (2, 1), (2, iii), (2, 5), (three, 2), (3, 4), (iv, one), (4, 3), (5, ii), (5, 6), (six, ane), (6, 5)] = fifteen

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

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

= 15/36
= 5/12

(xii) getting a doublet of even numbers:

Let E12 = effect of getting a doublet of fifty-fifty numbers. The events of a doublet of even numbers will be Eastward12 = [(2, ii), (four, 4), (6, 6)] = 3

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

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

= 3/36
= ane/12

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

Let E13 = consequence of getting a multiple of 2 on one dice and a multiple of iii on the other die. The events of a multiple of 2 on one die and a multiple of 3 on the other die will be Due eastthirteen = [(two, three), (2, vi), (3, 2), (iii, 4), (3, half-dozen), (4, 3), (4, six), (vi, ii), (6, iii), (6, 4), (6, vi)] = 11

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

               Number of favorable outcomes
P(Eastthirteen) = Full number of possible issue

= xi/36

4. Two dice are thrown. Find (i) the odds in favour of getting the sum five, and (2) the odds against getting the sum vi.

Solution:

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

Allow S be the sample space. Then, n(Due south) = 36.

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

Let Eastone be the result of getting the sum v. Then,
E1 = {(1, 4), (2, three), (3, 2), (4, 1)}
⇒ P(Eastwardi) = four
Therefore, P(E1) = n(Eastward1)/north(S) = 4/36 = one/9
⇒ odds in favour of E1 = P(E1)/[i – P(Due east1)] = (ane/9)/(1 – 1/9) = i/8.

(ii) the odds against getting the sum vi:

Let Due east2 exist the event of getting the sum 6. And then,
East2 = {(i, 5), (2, 4), (three, 3), (4, 2), (5, one)}
⇒ P(Etwo) = v
Therefore, P(Etwo) = due north(E2)/n(S) = five/36
⇒ odds against Due east2 = [1 – P(Eii)]/P(Etwo) = (1 – 5/36)/(five/36) = 31/5.

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

(i) equal numbers on both

(2) two numbers appearing on them whose sum is ix.

Solution:

The possible outcomes are

(1, 1), (1, ii), (i, iii), (1, four), (ane, 5), (1, 6),

(2, ane), (ii, ii), (2, 3), (2, iv), (2, 5), (2, half-dozen),

(3, 1), (3, 2), (3, 3), (3, four), (iii, 5), (three, half-dozen),

(four, one), (4, 2), (4, 3), (4, four), (iv, 5), (iv, half-dozen)

(five, i), (5, ii), (five, 3), (five, 4), (five, 5), (5, 6)

(6, 1), (half dozen, two), (half dozen, iii), (6, iv), (6, 5), (6, 6)

Sample Space of Rolling a Pair of Dice

Therefore, full number of possible outcomes = 36.

(i) Number of favourable outcomes for the upshot Eastward

                   = number of outcomes having equal numbers on both dice

                   = half-dozen    [namely, (1, 1), (2, two), (3, three), (iv, four), (v, 5), (6, vi)].

So, by definition, P(E) = \(\frac{6}{36}\)

                                 = \(\frac{i}{half dozen}\)

(two) Number of favourable outcomes for the result F

           = Number of outcomes in which two numbers actualization on them have the sum 9

            = iv     [namely, (3, 6), (iv, 5), (5, 4), (iii, 6)].

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

                                    = \(\frac{1}{ix}\).

These examples will assist us to solve different types of problems based on probability for rolling two dice.

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