Question

two different dice are tossed together find the probability (1) of getting a doublet and (11) of getting a sum of 10 of the numbers on the two dice?

Answer 1

Given that two different dice are tossed.

Total number of outcomes = 6^2

                                             = 36.


(1) Let A be the event of getting a doublet.

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

       = 6.

Required probability P(A) = 6/36

                                           = 1/6.


(ii) Let B be the event of getting a sum of 10 of the numbers of two dice.

n(B) = {6,4},{4,6},{5,5}

        = 3.


Required probability P(B) = n(B)/n(S)

                                          = 3/36

                                          = 1/12


Hope this helps!

Answer 2

Given:

Two dice

To find:

The probability of getting a double

The probability of getting a sum of 10

Solution:

Sample space = ( 1, 0 ), ( 1, 1 ), ( 1, 2 )... ( 6, 6 )

Hence,  

Sample space = 36

To find the probability of getting a doublet,

Outcomes = ( 1, 1 ), ( 2, 2 ).. ( 6, 6 )

Possible outcomes = 6

P (getting a double) = 6 / 36  

Hence, P (getting a double) = 1 / 6

To find the probability of getting a sum of 10,

Possible outcomes = ( 4, 6 ), ( 6, 4 ), ( 5, 5 )  

P (getting a sum of 10 ) =3/36

Hence, P ( getting a sum of 10 ) = 1 / 12

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