Question

3) Two dice are thrown. Find the probability of the
event given: The product of numbers on the
upper faces is 12
​

Answer 1

Answer:

0.0277777777777777777...

or 1/36

Step-by-step explanation:

the dice has faces of number 1 2 3 4 5 6 (6 faces)

so all probability are 6 × 6 = 36

and 12 can come only by 1 way that is 6,6

so the probability is

$\frac{1}{36}$

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Answer 2

Step-by-step explanation:

Total outcomes = (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)( 4,5 )(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 )

Possible event that the product of number on their upper face 12 are = (2,6),(3,4),(4,3),(6,2)

$\sf \boxed{P(E) = \frac{Total \: no \: of \: possible \: outcomes }{Toatl \: no \: of outcomes} }$

‎ ‎ ‎‎ ‎ ‎ ‎ ‎‎ ‎ ‎= 4\36 = 1\9

Therefore, The probability of the event that theprobability of the event that the product of numbers on their upper faces is 12 is 1\9