Question

In a single throw of two dice, find the probability of 1. A doublet 2.A number less than 3 on each dice 3.An even number as a sum 5.A total of atleast 10. Plz hlp me guys​

Answer 1

Answer:

i dont know

Step-by-step explanation:

Answer 2

1. 1/6

2.1/9

3.0

4.1/12

Step-by-step explanation:

As there are two dices

Total no of 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 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

Ist condition ( a doublet means when two dices show same no.) Now you can see here that no. of favourable outcomes=6 i.e. 1 1, 2 2, 3 3 , 4 4, 5 5, 6 6

P(E)=No. of fav outcomes/ total no. of outcomes

=6/36=1/6

2nd condition( there should be no.less than 3 on both sides)Now,here fav outcomes are: 1 1, 1 2, 2 1, 2 2

P(E)= No. of fav outcomes/Total no. of outcomes

= 4/36=1/9

3rd condition( an even no. should be there on both dice whose sum should be 5)Now,here fav outcomes are 0

P(E)=0

4th condition(the no. on both dices should make a total of 10 ) So here fav outcomes are:4 6, 5 5, 6 4

P(E)= 3/36

=1/12

Hope you understood.!!!!☺️