Suppose I have two types of light bulbs, type-ex and type-wye, whose lifetimes in years are given by exponential random variables \(X\) and \(Y\), such that \(\mathbb{P}(X >x) = e^{-x/2}\) for all \(x \geq 0\), and \(\mathbb{P}(Y >y) = e^{-y/3}\) for all \(y \geq 0\).
Suppose I install one type-ex light bulb and one type-wye light bulb, and the light bulbs function independently of one another. What is the probability that the ex light bulb dies before the wye light bulb?
Suppose I flip a fair coin 20 times. Assume that the coin flips are independent. Show that the probability that I get a consecutive run of at least four heads is greater than \(0.27\). Hint: group the coins into 5 blocks.
Suppose that when Tessa buys a Kinder surprise candy egg to get the Kinder surprise toy. Each egg contains one of \(n\) toys uniformly at random, independently of each other. She wishes to collect all \(n\) toys. Let \(T\) be the number of toys her father has to buy in order for her to collect all \(n\) toys. Compute \(\mathbb{E} T\). You should obtain a nice formula that will allow you to compute easily \(\mathbb{E} T\) if \(n=7\). Hint: let \(T_i\) be the number of toys that need to be purchased, after collecting \(i-1\) distinct toys, then \[ T = T_1 + \cdots +T_n.\]
Since \(X\) and \(Y\) are independent and their pdfs are given by \(f_X(x) = 1/2e^{-x/2}\) and \(f_Y(y) = 1/3e^{-y/3}\), when \(x,y \geq 0\), then the joint pdf is simply given by the product \(f_Xf_Y\); all the probabilities to do with \((X,Y)\) can be found by integration. Let \(R = \{(x,y): 0\leq x <y\}\). We compute
\[\begin{eqnarray*} \mathbb{P}(X <Y) &=& \int\int_R (\frac{1}{2})e^{-x/2}(\frac{1}{3})e^{-y/3}dxdy \\ &=& \int_0 ^{\infty} \frac{1}{2}e^{-x/2}\Big( \int_x^{\infty}\frac{1}{3}e^{-y/3}dy \Big) dx \\ &=& \int_0 ^{\infty} \frac{1}{2}e^{-x/2} \Big[ -e^{-y/3} \Big]_x^{\infty} dx \\ &=& \int_0 ^{\infty} \frac{1}{2}e^{-x/2}e^{-x/3} dx \\ %&=& \int_0 ^{\infty} \frac{1}{2}e^{-x/2}e^{-x/3} dx \\ &=& \int_0 ^{\infty} \frac{1}{2}e^{-5x/6} dx \\ &=& \Big[ -\frac{3}{5} e^{-5x/6} \Big]_0 ^{\infty} \\ &=& 3/5. \end{eqnarray*}\]
Let \(F\) be the event that there is a run of at least four heads. Partition the coin-flips into 5 blocks of 4. Consider the event \(E\) of getting four consecutive heads within at least one of the blocks. Clearly, \(E \subseteq F\). The probability of not getting four heads in a block is \(1-(1/2)^4\). So \(\mathbb{P}(E^c) =(1-(1/2)^4)^5\) and \(\mathbb{P}(E) = 1- \mathbb{P}(E^c) \approx 0.275803.\)
Note that \(T_1=1\), since at the start, any toy is new. For the next toy, any of the \(n-1\) remaining ones will do; if we get the one we first got, we have to try again. Thus the probability of success is \((n-1)/n\) and failure \(1/n\), and \(T_2\) is geometric with mean \(n/(n-1)\). Similarly, \(T_i\) is geometric with mean \(n/(n-i +1)\), and hence the linearity of expectation gives \[\begin{eqnarray*} \mathbb{E} T &=& \sum_{i=1} ^n \frac{n}{n-i+1} \\ &=& n \sum_{i=1}^n \frac{1}{n-i+1} \\ &=& n \sum_{i=1}^n \frac{1}{i}. \end{eqnarray*}\]