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3 dices are rolled simultaneously in how many outcomes will the facing side of the 2 dices shows the same number

plz explains your works and your reasoning ...

thanks

I am ready to be the first one to get it wrong again, mandy ...

Is that 30 ?

for the first dice, you can choose any of the 6 numbers,
for the second one, you can only one of 5 remaining numbers,
the the third one, you can only choose one - the same as the second dice.

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.