Given:
\[ |E|=72,\quad |M|=65,\quad |S|=50 \] \[ |E\cap M|=38,\quad |E\cap S|=30,\quad |M\cap S|=25,\quad |E\cap M\cap S|=15 \](a)
\[ |E\cup M\cup S| = 72+65+50-38-30-25+15=109 \] \[ P(E\cup M\cup S)=\frac{109}{120} \](b)
\[ 120-109=11 \]So 11 students take none of the three classes.
(c)
\[ |E\text{ only}|=72-38-30+15=19 \] \[ |M\text{ only}|=65-38-25+15=17 \] \[ |S\text{ only}|=50-30-25+15=10 \] \[ \text{Exactly one}=19+17+10=46 \](d)
\[ |E\cap M\text{ only}|=38-15=23 \] \[ |E\cap S\text{ only}|=30-15=15 \] \[ |M\cap S\text{ only}|=25-15=10 \] \[ \text{Exactly two}=23+15+10=48 \](e)
\[ P(E\mid M)=\frac{P(E\cap M)}{P(M)} = \frac{38/120}{65/120} = \frac{38}{65} \](f)
\[ P(E\cap M)=\frac{38}{120} \] \[ P(E)P(M)=\frac{72}{120}\cdot \frac{65}{120} \]These are not equal, so \(E\) and \(M\) are not independent.
(g)
No. Since \(E\) and \(M\) are not even pairwise independent, \(E,M,S\) cannot be mutually independent.
Total students:
\[ 8+7+6=21 \](a)
\[ \binom{21}{6}=54264 \](b)
\[ \binom{8}{2}\binom{7}{2}\binom{6}{2}=8820 \](c)
At least one from each major means the counts \((e,m,s)\) are positive and sum to 6. One way to write the answer is:
\[ \sum_{\substack{e+m+s=6\\ e,m,s\ge 1}} \binom{8}{e}\binom{7}{m}\binom{6}{s} \](d)
More Math majors than English majors:
\[ \sum_{\substack{e+m+s=6\\ m>e}} \binom{8}{e}\binom{7}{m}\binom{6}{s} \](e)
First choose the committee, then assign president, vice president, and secretary:
\[ \binom{21}{6}\cdot 6\cdot 5\cdot 4 \] \[ =54264\cdot 120=6511680 \](a)
\[ 36^8=2821109907456 \](b)
Choose positions for the digits, then fill digit positions and letter positions:
\[ \binom{8}{3}10^3 26^5=665357056000 \](c)
Use the complement: at least one digit = all passwords minus all-letter passwords.
\[ 36^8-26^8=2612282842880 \](d)
\[ 36\cdot 35\cdot 34\cdot 33\cdot 32\cdot 31\cdot 30\cdot 29 = 1220096908800 \](e)
Choose the digit positions, then choose distinct digits and distinct letters:
\[ \binom{8}{3}(10\cdot 9\cdot 8)(26\cdot25\cdot24\cdot23\cdot22) \] \[ =318269952000 \](a)
\[ x_1+x_2+x_3+x_4=12,\quad x_i\ge0 \] \[ \binom{12+4-1}{4-1}=\binom{15}{3}=455 \](b)
\[ x_1+x_2+x_3+x_4=12,\quad x_i\ge1 \] \[ \binom{12-1}{4-1}=\binom{11}{3}=165 \](c)
Let \(y_1=x_1-2\), \(y_2=x_2-1\), \(y_3=x_3\), and \(y_4=x_4\). Then:
\[ y_1+y_2+y_3+y_4=9,\quad y_i\ge0 \] \[ \binom{9+4-1}{4-1}=\binom{12}{3}=220 \](d)
Start with all nonnegative solutions, then subtract schedules where some day has at least 6 problems.
\[ \binom{15}{3}-4\binom{9}{3}+\binom{4}{2}\binom{3}{3} \] \[ =455-336+6=125 \]Total ways to choose 4 students:
\[ \binom{24}{4}=10626 \](a)
\[ P(\text{exactly 2 English}) = \frac{\binom{10}{2}\binom{14}{2}}{\binom{24}{4}} = \frac{4095}{10626} \](b)
\[ P(1E,1M,2S) = \frac{\binom{10}{1}\binom{8}{1}\binom{6}{2}}{\binom{24}{4}} = \frac{1200}{10626} \](c)
Since 4 students are chosen from 3 subjects, at least one from each subject means the count pattern is \(2,1,1\).
\[ P(\text{at least one from each}) = \frac{ \binom{10}{2}\binom{8}{1}\binom{6}{1} + \binom{10}{1}\binom{8}{2}\binom{6}{1} + \binom{10}{1}\binom{8}{1}\binom{6}{2} } {\binom{24}{4}} \] \[ = \frac{5040}{10626} \](d)
Given exactly 2 English students, the other 2 students are chosen from the 14 non-English students.
\[ P(\text{exactly 1 Math}\mid \text{exactly 2 English}) = \frac{\binom{8}{1}\binom{6}{1}}{\binom{14}{2}} = \frac{48}{91} \]There are 4 red balls and 8 non-red balls, with 12 balls total.
(a)
\[ P(X=k)=\frac{\binom{4}{k}\binom{8}{2-k}}{\binom{12}{2}}, \quad k=0,1,2 \](b)
While we didn't do hypergeometric distribution in class, the expected value for a hypergeometric distribution can be computed using linearity of expectation:
Say \(Y_i\) is the indicator random variable for drawing a red ball on the \(i\)-th draw, then \(X = Y_1 + Y_2\). By linearity of expectation: \[ E[X]=E[Y_1]+E[Y_2]=2\cdot \frac{4}{12}=\frac{2}{3} \](c)
To calculate the variance we will need to use the covariance: \(\mathrm{Cov}(Y_1,Y_2)=E[Y_1Y_2]-E[Y_1]E[Y_2]=P(Y_1=1,Y_2=1)-P(Y_1=1)P(Y_2=1)=4/12\cdot 3/11-1/9=-2/99 \) \[ \mathrm{Var}(X) = \mathrm{Var}(Y_1 +Y_2)=2Var(Y_1)+2\mathrm{Cov}(Y_1,Y_2)\\ =2\cdot \frac{4}{12}\cdot \frac{8}{12}+2\cdot\left(-\frac{2}{99}\right)\\ =\frac{4}{9}-\frac{4}{99}=\frac{40}{99} \](d)
No, \(X\) is not binomial because the draws are without replacement, so the trials are not independent and the success probability changes after each draw.
(a)
A student gets a question correct if they know it, or if they do not know it and guess correctly:
\[ q=0.7+0.3\cdot \frac14=0.775=\frac{31}{40} \](b)
\[ X\sim \mathrm{Binomial}\left(6,\frac{31}{40}\right) \](c)
\[ P(X=5) = \binom{6}{5} \left(\frac{31}{40}\right)^5 \left(\frac{9}{40}\right) \](d)
\[ E[X]=6\cdot \frac{31}{40}=\frac{93}{20}=4.65 \] \[ \mathrm{Var}(X)=6\cdot \frac{31}{40}\cdot \frac{9}{40} = \frac{837}{800} = 1.04625 \](a)
For \(k=0,1,2,3\):
\[ P(X=k)=\binom{3}{k}(0.5)^k(0.5)^{3-k} = \binom{3}{k}\left(\frac12\right)^3 \] \[ P(Y=k)=\binom{3}{k}\left(\frac12\right)^3 \](b)
\[ P(X=Y)=\sum_{k=0}^3 P(X=k,Y=k) \]By independence:
\[ P(X=Y)=\sum_{k=0}^3 P(X=k)P(Y=k) \] \[ = \sum_{k=0}^3 \left[\binom{3}{k}\left(\frac12\right)^3\right]^2 \] \[ = \frac{1^2+3^2+3^2+1^2}{64} = \frac{20}{64} = \frac{5}{16} \]For \(X\sim \mathrm{Geometric}(p)\), where \(X\) is the number of trials until the first success:
\[ P(X>r)=(1-p)^r \](a)
\[ P(X>m+n\mid X>m) = \frac{P(X>m+n)}{P(X>m)} \] \[ = \frac{(1-p)^{m+n}}{(1-p)^m} = (1-p)^n \](b)
\[ P(X>n)=(1-p)^n \]Therefore:
\[ P(X>m+n\mid X>m)=P(X>n) \](c)
Given that the first \(m\) trials failed, the expected additional waiting time is still \(1/p\). So:
\[ E[X\mid X>m]=m+\frac{1}{p} \](d)
Let \(Z=X-m\) be the additional number of trials needed after the first \(m\) failures. By memorylessness:
\[ Z\mid (X>m)\sim \mathrm{Geometric}(p) \]Thus:
\[ E[Z\mid X>m]=\frac{1}{p} \]Since \(X=m+Z\):
\[ E[X\mid X>m]=m+E[Z\mid X>m]=m+\frac{1}{p} \]