Question

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.


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Answer 1

Answer:

GIVEN :-

• 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

• Number of total outcome = n(S) = 36

TO FIND :-

▫️Find the probability of getting each sum from 2 to 9 separately.

FORMULAE :-

$P(E) = \frac{Number \: of \: favourable \: outcomes}{Total \: number \: of \: outcomes}$

SOLUTION :-

➡️(i) Let E1 be the event ‘getting sum 2’

Favourable outcomes for the event E1 = {(1,1),(1,1)}

n(E1) = 2

P(E1) = n(E1)/n(S) = 2/36 = 1/18

➡️(ii) Let E2 be the event ‘getting sum 3’

Favourable outcomes for the event E2 = {(1,2),(1,2),(2,1),(2,1)}

n(E2) = 4

P(E2) = n(E2)/n(S) = 4/36 = 1/9

➡️(iii) Let E3 be the event ‘getting sum 4’

Favourable outcomes for the event E3 = {(2,2)(2,2),(3,1),(3,1),(1,3),(1,3)}

n(E3) = 6

P(E3) = n(E3)/n(S) = 6/36 = 1/6

➡️(iv) Let E4 be the event ‘getting sum 5’

Favourable outcomes for the event E4 = {(2,3),(2,3),(4,1),(4,1),(3,2),(3,2)}

n(E4) = 6

P(E4) = n(E4)/n(S) = 6/36 = 1/6

➡️(v) Let E5 be the event ‘getting sum 6’

Favourable outcomes for the event E5 = {(3,3),(3,3),(4,2),(4,2),(5,1),(5,1)}

n(E5) = 6

P(E5) = n(E5)/n(S) = 6/36 = 1/6

➡️(vi) Let E6 be the event ‘getting sum 7’

Favourable outcomes for the event E6 = {(4,3),(4,3),(5,2),(5,2),(6,1),(6,1)}

n(E6) = 6

P(E6) = n(E6)/n(S) = 6/36 = 1/6

➡️(vii) Let E7 be the event ‘getting sum 8’

Favourable outcomes for the event E7 = {(5,3),(5,3),(6,2),(6,2)}

n(E7) = 4

P(E7) = n(E7)/n(S) = 4/36 = 1/9

➡️(viii) Let E8 be the event ‘getting sum 9’

Favourable outcomes for the event E8 = {(6,3),(6,3)}

n(E8) = 2

P(E8) = n(E8)/n(S) = 2/36 = 1/18