Question

Two dice are thrown simultaneously. Find the probability of getting: (2)

i) A multiple of 2 on one dice and a multiple of 3 on the other

ii) A doublet of even number
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Answer 1

Two dice are thrown simuntaneously

So all possible outcomes 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)

The total possible outcomes = 36

Now to find:

• i) A multiple of 2 on one dice and a multiple of 3 on the other

Solution,

Favourable outcomes = multiple of 2 on one dice and a multiple of 3 on the other

So,

(2,3) , (2,6) , (3,2) , (3,4) , (3,6) , (4,3) , (4,6) , (6,2) , (6,3) , (6,4) , (6,6)

Number of favourable outcome = 11

Now,

P(multiple of 2 on one dice and a multiple of 3 on the other ) = (Number of favourable outcomes/(Total number of possible outcomes)

= 11/36

Now

• ii) A doublet of even number

​Favorable outcomes for a doublet of even number are  = (2,2) , (4,4) , (6,6)

Number of favourable outcomes are = 3

Now,

P(doublet of even number) = (Number of favourable outcomes/(Total number of possible outcomes)

= 3/36 = 1/13

Hence,

• The probability of getting a multiple of 2 on one dice and a multiple of 3 on the other  is = 11/36

And

• The probability of getting a doublet of even number  = 1/12

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Answer 2

Answer:

i) A multiple of 2 on one dice and a multiple of 3 on the other = 11/36

ii) A doublet of even number = 1/13

Step-by-step explanation:

∴ Total number of elementary events = 6*6 = 36

The elementary events favorable are:  

(2, 3), (2, 6), (4,3), (4, 6), (6, 3), (6, 6), (3, 2), (3, 4), (3, 6), (6, 2), (6, 4)

Number of favorable outcomes = 11

P = 11/36

Now,

The doublet of even number are ( ​Favorable outcomes ) = (2,2) , (4,4) , (6,6)

The number of favorable outcomes are = 3

P = 3/36

Dividing by 3 to both the numerator and denominator

P = 1/13