Quiz 2 Solution

The following scenario applies to Questions 1 to 3.  Please provide numerical answers, accurate to at least 3 significant digits.

The lifetime $t$ of a device behaves according to the probability law $Pr([t,\mathrm{\infty }))=2^{-t/2}$ for $t\ge 0$.

1.  Find the probability of the event "lifetime is greater than or equal to 2 but less than 4."

Solution:  We wish to find $Pr([2,4))$.   Observing that $[2,\mathrm{\infty })=[2,4)\cup [4,\mathrm{\infty })$ is a union of disjoint events, we know from the Axiom III of probability that $Pr([2,\mathrm{\infty }))=Pr([2,4)]+Pr([4,\mathrm{\infty })$.  Thus

$Pr([2,4))=Pr([2,\mathrm{\infty }))-Pr([4,\mathrm{\infty }))$

$=2^{-2/2}-2^{-4/2}$

$=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}=0.25$

2.  Find the probability of the event "lifetime is less than 1."

Solution:  We wish to find $Pr([0,1))$.  Observing that the complementary event is $[1,\mathrm{\infty })$, we find

$Pr([0,1))=1-Pr([1,\mathrm{\infty }))=1-2^{-1/2}=1-\frac{1}{\sqrt{2}}\approx 0.292893$

3.  Find the probability of the event "lifetime is equal to 3."

Solution:  We wish to find $Pr([3,3])$. We can find this by computing the limit (for $ϵ>0$)

$Pr([3,3])=\underset{ϵ\to 0^{+}}{lim}Pr([3,3+ϵ))$

$=\underset{ϵ\to 0^{+}}{lim}(Pr([3,\mathrm{\infty }))-Pr([3+ϵ,\mathrm{\infty })))$

$=\underset{ϵ\to 0^{+}}{lim}(2^{-3/2}-2^{-(3+ϵ)/2})$

$=0$.

The following scenario applies to Questions 4 to 8.

A four-digit Personal Identification Number (PIN) is obtained by sampling the four digits with replacement and with ordering from an urn containing ten balls numbered  $\{0,1,\dots ,9\}$.

4.   What is the probability that the resulting four-digit PIN consists of the same digit repeated four times, i.e., that the PIN takes the form $aaaa$  for some digit $a$?

Solution:  The sample space contains ${10}^4$ outcomes, each of them having probability ${10}^{-4}$.  Since the digit $a$ can be chosen in 10 ways, there are exactly ten outcomes of the given form, namely $0000,1111,\dots ,9999$.  The probability that one of these is drawn is $10\cdot {10}^{-4}={10}^{-3}=0.001$.

5.  What is the probability that the resulting four-digit PIN consists of four distinct digits, i.e., a PIN with no repeated digits?

Solution:   The number of outcomes of the form $abcd$ where the digits are distinct  is equal to the number of possibilities that would be obtained when sampling without replacement, namely $P(10,4)=10\cdot 9\cdot 8\cdot 7=5040$.  However, we are sampling with replacement, so the probability of this event is $5040\cdot {10}^{-4}=0.504$.

6.  What is the probability that the four-digit PIN consists only of even digits or only of odd digits?  (Recall that the even digits comprise the set $\{0,2,4,6,8\}$ and the odd digits comprise the set $\{1,3,5,7,9\}$.  For example, 2468 and 3579 are PINs that consist only of even digits or only of odd digits.

Solution:   To obtain a PIN of the form EEEE (all even), we have 5 choices for the first digit, 5 choices for the second digit, etc., thus there are $5^4$ such PINs.  Similarly the number of PINs of the form OOOO (all odd) is also $5^4$.  These two subsets of PINs are disjoint, therefore

$Pr(\text{EEEE}\cup \text{OOOO})=Pr(\text{EEEE})+Pr(\text{OOOO})$

$=5^4\cdot {10}^{-4}+5^4\cdot {10}^{-4}$ $=2\cdot \frac{5^4}{{10}^4}$ $=2\cdot \frac{1}{2^4}$

$=\frac{1}{8}=0.125$.

7.  What is the probability that the resulting four-digit PIN consists of two even digits and two odd digits in any order?  (For example, 1234 and 2354 contain two even digits and two odd digits.)

Solution:  There are $(\genfrac{}{}{0ex}{}{4}{2})=6$ permutations containing two E's and two O's,  namely EEOO, EOEO, EOOE, OEEO, OEOE, OOEE.  These sets of PINs are disjoint and each have $5^4$ elements, there being 5 choices for each E and each O.   Thus the probability of the given event is

$6\cdot 5^4\cdot {10}^{-4}=\frac{3}{8}=0.375$

8.  What is the probability that the resulting four-digit PIN takes the form $aabb$ with $a,b\in \{0,1,\dots ,9\}$ and $a\ne b$?  For example, the PINs 1133 and 2277 are of this form.

Solution:  We have 10 choices for $a$ and 9 choices for $b$ once $a$ has been chosen, so there are $10\cdot 9=90$ PINs of this type.  The probability of occurrence is thus $90\cdot {10}^{-4}=9\cdot {10}^{-3}=0.009$.