Help need with Counting & Probability Question. Will MARK BRAINLEST IF ANSWER IS CORRECT. FIFTEEN points will be given!!!

Each of five, standard, six-sided dice is rolled once. Two of the dice come up the same, but the other three are all different from those two and different from each other. The pair is set aside, and the other three dice are re-rolled. The dice are said to show a "full house" if three of the dice show the same value and the other two show the same value (and potentially, but not necessarily, all five dice show the same value). What is the probability that after the second set of rolls, the dice show a full house?

Answer

PLZ MARK AS BRAINLIEST AND THX FOR THR BEST QUESTION

Of the three dice rolled, one may match the original pair and two are different, and this can occur three ways: ( 1/6 x 5/6 x 5/6 ) + ( 5/6 x 1/6 x 5/6 ) + ( 5/6 x 5/6 x 1/6 ) = 75/216

[ where 1/6 is the probability of matching the first two and 5/6 is the probability of not matching the first two ]

Or, two may match the original pair and one is different:

( 1/6 x 1/6 x 5/6 ) + ( 1/6 x 5/6 + 1/6 ) + ( 5/6 x 1/6 x 1/6 ) = 15/216

Or, all three may match the original pair: ( 1/6 x 1/6 x 1/6 ) = 1/216

Or, all three may be the same, but not match the original pair:

( 5/6 x 1/6 x 1/6 ) = 5/216

[ where 5/6 is the probability of getting a new number and 1/6 is the probability of matching that new number ]

Adding these together: 75/216 + 15/216 + 1/216 + 5/216 = 96/216 = 4/9

step-by-step explanation:

ok so im not 100% sure but you would start by adding 5.1t and 1.9t beacuse they have the same variable. that is 7t.   then add -3.6 + 1.2 which is -2.4

so then keep that as -2.4 + 7t. i apologize if this is incorrect. best luck on your test < 3

step-by-step explanation: