1) A balanced die is tosssed six times, and the number on the uppermost face is recorded each time. What is the probability that the numbers recorded are 1, 2, 3, 4, 5, and 6 in any order? My attempted answer is (6!)/(6^6), is it correct? 2) Suppose a die has been altered so that the faces are 1, 2, 3, 4, 5, and 5. If the die is tossed five times, what is the probability that the numbers recorded are 1, 2, 3, 4, and 5 in any oder? No clue...can anyone help me, please? 3) We need to arrange 5 math books, 4 physics books and 2 statistics books on a shelf. How many possible arrangements exists so that books of the same subjects will lie side by side? Is it 3! x (5! x 4! x 2!)? Thanks for helping!
1. P = (6/6)(5/6)(4/6)(3/6)(2/6)(1/6) because the number of allowable digits goes down as you make more rolls. (No number seen can reappear.) Thus \(P = {{6 !}\over{6^6}} = {{5 !}\over{6^5}} = \frac{120}{7776} = \frac{5}{324} \) 2. \(P = P_1 + P_2 + P_3 + P_4 + P_5\) where \(P_n\) is that the only once allowed 5 comes up in the nth position. We have: \(P_1 = \frac{2}{6} \times \frac{4}{6} \times \frac{3}{6} \times \frac{2}{6} \times \frac{1}{6} \) \(P_2 = \frac{4}{6} \times \frac{2}{6} \times \frac{3}{6} \times \frac{2}{6} \times \frac{1}{6} \) \(P_3 = \frac{4}{6} \times \frac{3}{6} \times \frac{2}{6} \times \frac{2}{6} \times \frac{1}{6} \) \(P_4 = \frac{4}{6} \times \frac{3}{6} \times \frac{2}{6} \times \frac{2}{6} \times \frac{1}{6} \) \(P_5 = \frac{4}{6} \times \frac{3}{6} \times \frac{2}{6} \times \frac{1}{6} \times \frac{2}{6}\) Which are all the same, thus \(P = 5 P_1 = 10{{4 !}\over{6^5}}= \frac{240}{7776} = \frac{5}{162}\) 3. You have correctly computed the ways of arranging the books so that all the math, physics and statistics are in uninterrupted blocks. I think that is what the question is asking for.
You might find it easier to see the answer to the second question as: \( \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \times \frac{2}{6} \times 5! = \frac{5}{162} \)
