Math, asked by minalpawar745, 3 months ago

SzSR, R2, R
Q 3A Complete the given activity.
(1) Two dice are thrown, find the probability of the following event.
Event D : The sum of the numbers on upper faces is at least 10.
Solution : Sample space - 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)}
(S)
Event D: The sum of the numbers on upper faces is at least 10.

Answers

Answered by hemntmali2001
2

Step-by-step explanation:

Two dice are thrown.

Sample space = 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

3) Let E be the event that the sum of

the numbers on their upper face is at

least 8.

E = {(2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (4,

6), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6,

3), (6, 4), (6,5), (6, 6)}

So, n(E) = 15

Hence, Probability = nCE

)

= 15

=

5

12

36

1) Let F be the event that the sum of

the numbers on their upper face is at

most 4.

F = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1)}

So, n(F) = 6

Hence, Probability = n(F)

)

=

6

36

= 1

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