Math, asked by shrekush31, 1 month ago

.Two dice are thrown .find the probability of getting (a).A sum of 10 (b). a multiple of 3 as the sum (c).an even number on the first dice and an odd number on the second dice.​

Answers

Answered by karanseervi360
0

Step-by-step explanation:

Total number of outcomes =6

2

=36

1) Cases favourable to an even nos are the sum

(1,1),(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5)(4,2),(4,4),(4,6),(5,1),(5,3),(5,5),(6,2),(6,4),(6,6)

=18cases

2) Cases favourable sum as prime nos

3(1,1),(1,2),(1,4),(1,6),(2,1)(2,3),(2,5),(3,2),(4,1),(5,2),(5,6),(6,1)(6,5)

No of cases =13

Required probability =13

3) Cases favourable to total of at least 10

(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)

No. of cases =6

Required of probability

36

6

=

6

1

4) Favourable case of doublet of even nos.

=(2,2),(4,4),(6,6)

No. of cases=3

Required probability =

36

3

=

12

1

5) Favourable cases as a multiple of 3 as the sum =(1,2),(1,5),(2,4),(2,6)

=(3,3),(3,6),(4,2),(4,8)

=(5,1),(5,4),(6,3),(6,6)

No. of cases =12

Required probability =

36

12

=

3

1

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