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In a simultaneous throw of a pair of dice,

find the probability of getting:

(i) 8 as the sum

(ii) a doublet

(iii) a doublet of prime numbers

(iv) a doublet of odd numbers

(v) a sum greater than 9

(vi) an even number on first

(vii) an even number on one and a multiple of 3 on the other

(viii) neither 9 nor 11 as the sum of the numbers on the faces

(ix) a sum less than 6

(x) a sum less than 7

(xi) a sum more than 7

(xii) at least once

(xiii) a number other than 5 on any dice.

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In a throw of pair of dice, total no of possible outcomes = 36 (6 × 6) which are

(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)

(i) Let E be event of getting the sum as 8

No. of favorable outcomes = 5 {(2, 6) , (3, 5) , (4, 4) , (5, 3) , (6, 2)}

We know that, Probability P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes)

P(E) = 5/36

(ii) E ⟶ event of getting a doublet

No. of favorable outcomes = 5 {(1, 1) , (2, 2) , (3, 3) , (4, 4) , (5, 5) , (6, 6)}

Total no. of possible outcomes = 36

P(E) = 6/36 = 1/6

(iii) E ⟶ event of getting a doublet of prime no’s

No. of favorable outcomes = 3 {(2, 2) , (3, 3) , (5, 5)}

Total no. of possible outcomes = 36

P(E) = 3/36 = 1/12

(iv) E ⟶ event of getting a doublet of odd no’s

No. of favorable outcomes = 3 {(1, 1) , (3, 3) , (5, 5)}

Total no. of possible outcomes = 36

P(E) = 3/36 = 1/12

(v) E ⟶ event of getting a sum greater than 9

No. of favorable outcomes = 6 {(4, 6) , (5, 5) , (5, 6) , (6, 4) , (6, 5) , (6, 6)}

Total no. of possible outcomes = 36

P(E) = 6/36 = 1/6

(vi) E ⟶ event of getting an even no. on first

No. of favorable outcomes = 18 {(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)}

Total no. of possible outcomes = 36

P(E) = 18/36 = 1/2

(vii) E ⟶ event of getting an even no. on one and a multiple of 3 on other

No. of favorable outcomes = 11 {(2, 3) , (2, 6) , (4, 3) , (4, 6) , (6, 3) , (6, 6) , (3, 2) , (3, 4) , (3, 4) , (3, 6) , (6, 2) , (6, 4)}

Total no. of possible outcomes = 36

P(E) = 11/36

(viii) Bar E ⟶ event of getting neither 9 nor 11 as the sum of numbers on faces

E ⟶ getting either 9 or 11 as the sum of no’s on faces

No. of favorable outcomes = 6 {(3, 6) , (4, 5) , (5, 4) , (6, 3) , (5, 6) , (6, 5)}

Total no. of possible outcomes = 36

P(E) = 6/36 = 1/6

P(Bar E) = 1 - P = 1 - 1/6 = 5/6

(ix) E ⟶ event of getting a sum less than 6

No. of favorable outcomes = 10 {(1, 1) , (1, 2) , (1, 3) , (1, 4) , (2, 1) , (2, 2) , (2, 3) , (3, 1) , (3, 2) , (4, 1)}

Total no. of possible outcomes = 36

P(E) = 10/36 = 5/18

(x) E ⟶ event of getting a sum less than 7

No. of favorable outcomes = 15 {(1, 1) , (1, 2) , (1, 3) , (1, 4) , (1, 5) , (2, 1) , (2, 2) , (2, 3) , (2, 4) , (3, 1) , (3, 2) , (3, 3) , (4, 1) , (4, 2) , (5, 1)}

Total no. of possible outcomes = 36

P(E) = 15/36 = 5/12

(xi) E ⟶ event of getting a sum more than 7

No. of favorable outcomes = 15 {(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)}

Total no. of possible outcomes = 36

P(E) = 15/36 = 5/12

(xii) E ⟶ event of getting a 1 at least once

No. of favorable outcomes = 11 {(1, 1) , (1, 2) , (1, 3) , (1, 4) , (1, 5) , (1, 6) , (2, 1) , (3, 1) , (4, 1) , (5, 1) , (6, 1)}

Total no. of possible outcomes = 36

P(E) = 11/36

(xiii) E ⟶ event of getting a no other than 5 on any dice

No. of favourable outcomes = 25 {(1, 1) , (1, 2) , (1, 3) , (1, 4) , (1, 6) , (2, 1) , (2, 2) , (2, 3) , (2, 4) , (2, 6) , (3, 1) , (3, 2) , (3, 3) , (3, 4) , (3, 6) , (4, 1) , (4, 2) , (4, 3) , (4, 4) , (4, 6) , (6, 1) , (6, 2) , (6, 3) , (6, 4) , (6, 6)}

Total no. of possible outcomes = 36

P(E) = 25/36

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