Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is

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27 Dec 2016 Probability

Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is greater than 15?

OPtion

1) 7/36
2) 8/36
3) 9/36
4) 5/36
5) 10/36
6) 11/36
7) 1/3
8) 13/36
9) 14/36
10)15/36

Solution

Total Number of possible outcomes = 6C1*6C1 = 36
Positive outcomes must satisfy D1*D2> 15
For D1 or D2 = 1 and 2 there is no possibility of getting D1*D2 > 15
For D1 = 3, we have D2=5 and 6 which satisfies D1*D2 > 15
(3,5) and (3,6)
For D1 = 4, we have D2=4, 5 and 6 which satisfies D1*D2 > 15
(4,4), (4,5) and (4,6)
For D1 = 5, we have D2=4, 5 and 6 which satisfies D1*D2 > 15
(5,4), (5,5) and (5,6)
For D1 = 6, we have D2=3, 4, 5 and 6 which satisfies D1*D2 > 15
(6,3), (6,4), (6,5) and (6,6)

Count of positive outcome = 11
Probability = Positive outcome/Total Possible Outcomes = 11/36
Option 6)

Correct Option: 6 Best Solution (18)

G.S.Kantharao   8 years AGO

Total outcomes = 36
Required outcomes= 11
(3,6),(4,4),(4,5),(4,6),(5,4),(5,5),(5,6),(6,3),(6,4),(6,5),(6,6) = 11
Hence probability = 11/36.
Hence the right option is 6.

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abhishek   8 years AGO

total cases: 36
possible cases:(3,6),(4,4),(4,5),(4,6),(5,4),(5,5),(5,6),(6,3),(6,4),(6,5),(6,6)
therefore probabilty: 11/36 option:6)

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Divya Vasudevan   8 years AGO

Total number of events n(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)}=36
When their produt is greater than 15 n(E)=(3,6)(4,5)(4,6)(4,4)(5,6)(5,4)(5,5)(6,3)(6,4)(6,5)(6,6)=11
P(E)=n(E)/n(S)=11/36

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Umesh sahu   8 years AGO

possible combination is (6,3) ,(3,6),(6,6),(6,4),(6,5),(5,6),(4,6)(4,5)(5,4),(4,4)(5,5)

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JENSI KARUNYA.S   8 years AGO

S={(1,1) ,(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),........,(6,6)}
probability of getting the product of two numbers is greater than 15=11/36
option 6) 11/36 is correct answer.

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kamaniharitha   8 years AGO

I calculated result in reverse order.(i.e probability of getting two number product less than or equal to 15)
with 1 : (1*1),(1*2),(1*3),(1*4),(1*5),(1*6) here we have 6 possibilites similarly
with 2 we have 6 ,with 3 -5 ,with 4 - 3 ,with 5 -3 and with 6-2
add all the possibilites we get 25
requried probability = 1-(25/36)
so answer is 11/36

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Vivek Yadav   8 years AGO

Total case = 6*6 = 36
first dice outcomes -> second dice outcomes accordingly:
1 : No result from second dice will result 15 on multiply
2 : No result from second dice will result 15 on multiply
3 : 3*6 Result = 1
4 : 4*4, 4*5, 4*6 result = 3
5 : 5*4, 5*5, 5*6 Result = 3
6 : 6*3, 6*4, 6*5, 6*6 Result = 4

Total result according to question 1 + 3 + 3 + 4 = 11

Hence answer = 11/36

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Athulya   8 years AGO

Sample space S=36
possible outcomes R=(3,6) (6,3),(4,4),(4,5),(5,4),(4,6),(6,4),(5,5),(5,6),(6,5),(6,6)

Ans: P(R)=n(R)/n(S)=11/36

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Venkat   8 years AGO

two dices are thrown simultaneously
no.of possible ways=(3,6)(4,4)(4,5)(4,6)(5,4)(5,5)(5,6)(6,3)(6,4)(6,5)(6,6)
total possible ways=6^2=36
so getting probability of product 15=11/36

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Shivani Pawar   8 years AGO

possible outcomes are (3,6)(4,4)(4,5)(4,6)(5,4)(5,5)(5,6)(6,3)(6,4)(6,5)(6,6)

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Jebas Gnana Prakash E   8 years AGO

Multiply first dice{1,2,3,4,5,6} to second dice{1,2,3,4,5,6} and count the numbers got 11
So the ans is 11/36

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lingadurai   8 years AGO

Two dice are 1/36
two numbers are greater then 15
That the probability =11/36

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Abhishek Kumar   8 years AGO

(3,6)(4,4)(4,5)(4,6)(5,4)(5,5)(5,6)(6,3)(6,4)(6,5)(6,6)
=11
answer=11/36

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Mathi Vanan P   8 years AGO

Multiply each no in first dice{1,2,3,4,5,6} to the second dice{1,2,3,4,5,6} and count the result got the ans
So the ans is 11/36

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VISHAL.MORI   8 years AGO

Product is more than 15 is for...
6*6 (1)
6*5 (2)
6*4 (2)
6*3 (2)
5*5 (1)
5*4 (2)
4*4 (1)

so, 11/36

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vivek   8 years AGO

product of number >15 are- (6,6),(6 5),(6 4),(6 3) ,(5,6),(5 5),(5,4) (4 6),(4,5),(4,4) (3,6)

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PremaDeepika   8 years AGO

total combinations=36
combinations greater than 15 are11
so 11/36

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Mukund Manohar   8 years AGO

Required Probability = Positive Outcome/Total Possible Outcome
= 11/36

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