Question

b) If two dice are thrown together, find the probability of getting:
(i) the sum as a prime number.
(ii) a doublet.​

Answer 1

Answer:

(i) 7/18

(ii) 1/6

Step-by-step explanation:

(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)

:. n (s) = 6×6

= 36

(i) let A be the event that the sum of the uppermost faces are prime number

A = {(1,2) (1,4) (1,6) (2,1) (2,3) (2,5) (3,2) (3,4) (4,1) (4,3) (5,2) (5,6) (6,1) (6,5)}

n (A) = 14

:. Probability of getting 'A' = n (A) / n (S)

14/36

7/18 ans.

(ii) let B be the event that there are doublet of number.

B = {( 1,1) (2,2) (3,3) (4,4) (5,5) (6,6)}

:. Probability of getting 'B' = n (B) /n (S)

6/36

1/6 ans.

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