blue die a yellow die and a white die are thrown at the same time. What is the probability that the sum of three numbers that turn up is 10 ?
Answers
Answer:
In a throw of pair of dice - white and black, total no of possible outcomes=36(6×6) which are
(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)
Solution (i):
Let E be event of getting the sum as 8
No. of favorable outcomes =5 (i.e.,(2,6),(3,5),(4,4),(5,3),(6,2))
We know that, Probability P(E) =
(Total no.of possible outcomes)
(No.of favorable outcomes)
=
36
5
Solution (ii):
Let E be event of obtaining the total 6
No. of favorable outcomes =5 (i.e.,(1,5),(2,4),(3,3),(4,2),(5,1))
We know that, Probability P(E) =
(Total no.of possible outcomes)
(No.of favorable outcomes)
=
36
5
Solution (iii):
Let E be event of obtaining the total 10
No. of favorable outcomes =3 (i.e.,(4,6),(5,5),(6,4))
We know that, Probability P(E) =
(Total no.of possible outcomes)
(No.of favorable outcomes)
=
36
3
=
12
1
Solution (iv):
Let E be event of getting the same number on both dice
No. of favorable outcomes =6 (i.e.,(1,1),(2,2),(3,3),(4,4),(5,5),(6,6))
We know that, Probability P(E) =
(Total no.of possible outcomes)
(No.of favorable outcomes)
=
36
6
=
6
1
Solution (v):
E⟶ event of obtaining the total more than 9
No. of favorable outcomes =6 (i.e.,(4,6),(5,5),(5,6),(6,4),(6,5),(6,6))
P(E)=
36
6
=
6
1
Solution (vi):
E⟶ event of obtaining the sum of the two numbers appearing on the top of the dice as 13
No. of favorable outcomes =0 (i.e.,no combination of outcomes add up to 13. The maximum sum that can be obtained is 12)
P(E)=0
Solution (vii):
E⟶ event of obtaining the sum of the two numbers appearing on the top of the dice as less than or equal to 12
No. of favorable outcomes =36 =Total No. of favorable outcomes
(i.e.,(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))
P(E)=
36
36
=1
Solution (viii):
Let E be event of getting the product of numbers appearing on the top of the dice less than 9
No. of favorable outcomes =16 (i.e.,(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)(2,1)(2,2)(2,3)(2,4)(3,1)(3,2)(4,1)(4,2)(5,1)(6,1))
We know that, Probability P(E) =
(Total no.of possible outcomes)
(No.of favorable outcomes)
=
36
16
=
9
4
Solution (ix):
Let E be event of getting the difference of the numbers appearing on the top of the two dice as 2
No. of favorable outcomes =8 (i.e.,(1,3)(2,4)(3,1)(3,5)(4,2)(4,6)(5,3)(6,4))
We know that, Probability P(E) =
(Total no.of possible outcomes)
(No.of favorable outcomes)
=
36
8
=
9
2
Answer:
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