Ques:- Three 6 faced dice are thrown together. The probability that exactly two dice show the same number on them is -.
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Recent Answer : :
7days
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A. 20
B. 21
C. 22
D. 24
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A. 15
B. 20
C. 30
D. 35
Recent Answer : :
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Ques:- Which greatest number will divide 3026 and 5053 leaving remainders 11 and 13 respectively?
A. 15
B. 30
C. 40
D. 45
Recent Answer : :
Answer is 45
First we need to subtract those reminders from the respective numbers, then we have to find the hcf of two numbers(numbers got from the subtraction) then you will get the answer.
So,
After subtraction you will get
3026-11 = 3015
5053-13 = 5040
HCF of these two numbers
5 | 3015 5040
3 | 603 1008
3 | 201 336
| 67 112
We can’t find a common diviser since 67 is a prime number
So the HCF = 5 * 3 * 3
= 45
5*3*3 = 45
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Let A, B and C be the three 6-faced dice.
Then, according to the question,
Since two dices has to be equal, that value can be any of the 6 faces, i.e., 6C1 cases.
Now for each case, 2 equal dices can be selected from 3 dices in 3C2 i.e., 3 ways.
And for each of the above, the third dice can have any of the 5 remaining faces
The possible outcomes are P(A)=61,P(B)=61,P(C)=65,P(A)=61,P(B)=65,P(C)=61 and P(A)=65,P(B)=61,P(C)=61
Hence the required probability = 61×61×65×6×3=21690=125