Question

Two dice are throw at the same time write down all the possible outcomes.
What is the probability that the sum of the tow numbers appearing on the
top of the dice is
(i) 8? (2) 13.​

Answer 1

Answer:

hey mate, ...

Step-by-step explanation:

If two dice are thrown ...

so, let 'S' be the sample space..

therefore,

S - {(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)}

So, n(S) - (36)

(¡) Sum of the two number appearing on the dice is 8.

let 'A' be the event ...

A -{(2,6),(3,5),(4,4),(5,3),(6,2)}

n(A) - 5 ..

P(A) = n(A)/ n(S) = 5/36.

(ii) Sum of the two number appearing on the upper surface of dice is. 13.

Let, 'B" be the event..

B - {()}... as sum of no number here, is 12..

n(B) - 0..

P(B) - 0 ... as anything divided by 0 is 0....

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