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Re: Ps Rolled dices [#permalink]
OA is 90 I will post the OA in a few days if needed
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Re: Ps Rolled dices [#permalink]
mandy wrote:
OA is 90 I will post the OA in a few days if needed


Now that you gave us the answer I know the correct approach

total outcomes = 6x6x6 = 216
outcomes with all distinct outcomes = 6x5x4 = 120
outcomes with all same numbers on the 3 dices = 6

216 - 120 - 6 = 90

good one thank you
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Re: Ps Rolled dices [#permalink]
This is a nice question.

Total ways = 6* 6* 6 = 216

Ways when all three have same numbers = 6
Ways when none of them has same numbers = 6* 5 * 4 = 120

ANS = 216-6-120 = 90
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Re: Ps Rolled dices [#permalink]
ps_dahiya wrote:
This is a nice question.

Total ways = 6* 6* 6 = 216

Ways when all three have same numbers = 6
Ways when none of them has same numbers = 6* 5 * 4 = 120

ANS = 216-6-120 = 90

Good that we have you here)))Gr8 explanation
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Re: Ps Rolled dices [#permalink]
First, calculate the probability that a given number ( say 1 ) comes up on two of the dices

1/6*1/6*5/6 ( 5/6 since the third number can be anything but 1 ).

Since there are 6 numbers we multiply it by 6.

The result is (1/6*1/6*5/6)*6. Since for any given outcome with 2 dice having the same number, the outcome can be arranged in 3!/2! = 3 ways, We need to multiply that as well,
(1/6*1/6*5/6)*6*3 = 90/216 is the probability. Thus, the total favourable outcomes are 90.

BTW, I like p_s_dahiya's method better. No need to involve prob...just count.



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