Question

A cubical die numbered from 1 to 6 is rolled twice. Find the probability of
getting the sum of numbers on its faces is 10.
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Answer 1

HEY MATE YOUR ANSWER IS HERE

EVENT: A dice is rolled twice.

$sample \: space \: =$

{(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) }

Therefore,

$n(s) = 36$

Let A be the event where sum of the number on its faces is 10

Hence,

A ={(4,6),(5,5),(6,4)}

$therefore \\ n(a) = 3$

$probability \: of \: a = \binom{n(a)}{n(s)} \\ \\ probability \: of \: a \: = \binom{3}{36} \\ \\ therefore \: \\ probability \: of \: a = \binom{1}{12}$

Hence, probability of getting a sum of 10= 1/12


HOPE IT HELPS YOU

Answer 2

Answer:

1/12 is your answer

Step-by-step explanation:

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