Selected Answers to Problem Set 2

---

4. (a) 76 base 10 = 64 + 8 + 4 = 2⁶ + 2³ + 2² = 100110 base 2
(b) 4321 base 8 = 4(8³) + 3(8²) + 2(8¹) + 1(8⁰) = 2048 + 192 + 16 + 1 = 2257

---

5. We want to find the probability of the first player or the second player or the third player getting a hit. This is equivalent to the complement of the first player and the second player and the third player not getting a hit.

Someone getting a hit = 1 - (1 - 0.3)(1 - 0.2)(1 - 0.25)
                      = 1 - (0.7)(0.8)(0.75)
                      = 1 - 0.42
                      = 0.58

---

6. We want to find the probability that bus 1 arrives before t minutes or bus 2 arrives before t minutes. This is equivalent to the complement that both bus 1 and bus 2 do not arrive before t minutes.

A bus arrives before time t = 1 - (1 - t/8)(1 - t/10)
                            = 1 - (t2/40 - (9/40)t + 1)
			    = -t2/40 - (9/40)t

This function gives us the probability that a bus will arrive before t minutes. We want to know the probability that a bus arrives at exactly t minutes. We take the derivative to achieve this end. We use the derivative with the density function over the integral of all possible values of t, from 0 to 8, to get 2.93 minutes.

We find the time for three buses in a similar fashion.

A bus arrives before time t = 1 - (1 - t/10)3
                            = 1 - (1 - (t/10) + (t2/100) - t3/1000)
			    = (t/10) - (t2/100) + (t3/1000)

Again, we take the integral over all possible values of t on the density function (the derivative of our previous equation), from 0 to 10, to get 2.5 minutes.

---