A fair die is thrown 3 times what will be the probability of getting 3 consecutive numbers in inc

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29 Nov 2010 die thrown

A fair die is thrown 3 times what will be the probability of getting 3 consecutive numbers in increasing order.

OPtion

1) 1/2
2) 1/3
3) 1/12
4) 1/6
5) 1/216
6) 1/72
7) 1/144
8) 1/36
9) 1/54
10) 1/108

Solution

there are 4 favorable case
(1,2,3)
(2,3,4)
(3,4,5)
(4,5,6)
so the probability is 1/6*1/6*1/6*4
=4/216
=1/54

Correct Option: 9 Best Solution (5)

Revathi   13 years AGO

there are 4 consecutive numbers in inc order
1,2,3
2,3,4
3,4,5
4,5,6
therefore probability is 4/216
=1/54

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madhuri   13 years AGO

there are 4 cases to get consecutive numbers in inc order
1,2,3 2,3,4 3,4,5 4,5,6
probability=4/(6*6*6)
=1/54

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Dr.Dipin Singh   13 years AGO

A fair die is thrown 3 times . for getting 3 consecutive numbers in increasing order, first number should be 1,2,3,4 out of 6 possible numbers.

So favourable possibility is 4/6.
For getting next consecutive number in next throw, probability is 1/6.
For getting next consecutive number in next throw, probability is again 1/6.
Hence probability of getting 3 consecutive numbers in increasing order = (4/6)*(1/6)*(1/6)=4/216 =1/54.

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VIJI   13 years AGO

Sample space =216
Getting 3 consecutive no in increasing order are(1,2,3)(2,3,4)(3,4,5)(4,5,6)
probability=4/216=1/54

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SUDESHNA SEN   13 years AGO

Total no. of possible outcomes= 6^3=216.
3 consecutive numbers in increasing order can come in the following 4 ways: {(1,2,3),(2,3,4),(3,4,5),(4,5,6)}.
So, required probability= (4/216)= (1/54)..... Option 9)

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