An unbiased usual six-sided die is thrown three times. The sum of the numbers coming up is 10. The probability that 2 has appeared at least once is
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If a dice is rolled 3 times,
there are 6³ = 216 permutations of numbers obtained. So,
In order to get sum of all 3 numbers as 10, we can write possible combinations first,
There are 6 different combinations where 3 die can add up to 10:
Firstly, 1-3-6, 1-4-5, 2-3-5
Each of the following has 6 permutations as we can arrange each in 6 ways
Secondly, 2-2-6, 2-4-4, 3-3-4
Each of the following has 3 permutations as we can arrange each in 3 ways
Our 2nd option only has digit 2 in it,
so, no. of ways we will have 2 atleast once will be = 6
because in 2 such pairs digit 2 is occurring, and each pair is achieved in 3 ways
Probability of having 2 at least once in such situation = 6/216 = 1/36
there are 6³ = 216 permutations of numbers obtained. So,
In order to get sum of all 3 numbers as 10, we can write possible combinations first,
There are 6 different combinations where 3 die can add up to 10:
Firstly, 1-3-6, 1-4-5, 2-3-5
Each of the following has 6 permutations as we can arrange each in 6 ways
Secondly, 2-2-6, 2-4-4, 3-3-4
Each of the following has 3 permutations as we can arrange each in 3 ways
Our 2nd option only has digit 2 in it,
so, no. of ways we will have 2 atleast once will be = 6
because in 2 such pairs digit 2 is occurring, and each pair is achieved in 3 ways
Probability of having 2 at least once in such situation = 6/216 = 1/36
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