Question

Five standard six-sided dice are rolled. We are told there is no three-of-a-kind, but there is a pair of dice that show the same number. These two dice are set aside, and the other three dice are re-rolled. What is the probability that after re-rolling these three dice, at least three of the five dice show the same value

Answer 1

The probability that after re-rolling these three dice, at least three of the five dice show the same value is  $\frac{4}{9}$

Of the three dice rolled, one may match the original pair and two are different, and this can occur three ways:

$(\frac16 \times \frac56 \times \frac56 ) + ( \frac56 \times \frac16 \times \frac56 ) + ( \frac56 \times \frac 56 \times \frac 16 ) = \frac {75}{216}$

[ where $\frac{1}{6}$ is the probability of matching the first two and $\frac{5}{6}$ is the probability of not matching the first two ]

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

$( \frac16 \times \frac16 \times \frac 56 ) + ( \frac16 \times \frac56 \times \frac 16 ) + ( \frac56 \times \frac16 \times \frac16 ) = \frac {15}{216}$

Or, all three may match the original pair

$( \frac 16 \times \frac 16 \times \frac 16 ) = \frac {1}{216}$

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

$( \frac 56 \times \frac 16 \times\frac 16 ) = \frac {5}{216}$

[ where $\frac{5}{6}$ is the probability of getting a new number and $\frac{1}{6}$ is the probability of matching that new number ]

Adding these together: $\frac {75}{216} + \frac {15}{216} + \frac 1{216} + \frac 5{216} = \frac {96}{216} = \frac 49$

To learn more about probability with the given link

brainly.com/question/11234923

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