M4maths Previous Puzzles

03 Jul 2013 Probability

What are the odds in favour of throwing atleast 7 in a single throw of two dice.

OPtion

1) 3/2
2) 1/3
3) 7/5
4) 6/5
5) 2/5
6) 5/2
7) 4/3
8) 7/4
9) 5/7
10) None of these

Solution

Total number of possible outcomes = 36 (solve)
Let E be the event "Sum of the numbers on the two dice is atleast 7
therefore
E = (1,6)(2,5)(3,4)(4,3)(5,2)(6,1)(2,6)(3,5)(4,4)(5,3)(6,2)(3,6)(4,5)(5,4)(6,3)(4,6)(5,5)(6,4)(5,6)(6,5)(6,6)
n(E)=21, n (not E) = 36-21=15
P(E)=21/36 and P(not E) = 15/36
Odd is favour of the event E = P(E):p(not E)
(21/36):(15/36) = 7:5
7/5
option 3

Correct Option: 3 Best Solution (1)

Sneha Agarwal   11 years AGO

odds in favour = favourable outcomes/unfavourable outcomes
unfavourable outcomes = 15
odd is favour = 21/15
= 7/5

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