Question

three dices are thrown simultaneously. find the probability that two of them show the same face

Answer 1

Since you need exactly two to be the same, there are three possibilities: 1. First and second, not third 2. First and third, not second 3. Second and third, not first

For 1) The first die, you have 6/6. The second die needs to be equal to the first, so you have probability of 1/6. Then the third die can't be equal to the first and second dice, so it's 5/6. All together you get 1× 1/6 × 5/6 And since the next two cases yield the same results, then the probability that the same number apears on exactly two of the three dice is

3× (1 × 1/6 × 5/6 ) =5/12

Did I do this correctly? Thank you.