Учебные пробникиPractice mocks

$p=2/5$. For exactly $k$ alarms, choose their positions and multiply the independent outcome probabilities: $P(N=k)=\binom{6}k p^k(1-p)^{6-k}$. Sum over $k\in\{0,1\}$: $P=0.2333$. Each count is a disjoint case; this is not a sum of overlapping “at least” events.

A favorable subset combines choices from three groups: $\binom{4}{2}\binom{3}{1}\binom{3}{1}$. There are $\binom{10}{4}$ subsets in total. Divide to get $P=0.2571$. No extra order factor is needed: both counts refer to subsets.

First update the source. The probability of the first white is $z=3/10\cdot2/5+(1-3/10)\cdot4/5=17/25$. After observing it, the weights of A and B are $3/17$ and $14/17$: divide each corresponding contribution by $z$. They sum to $1$.

The required value is the second weight: $14/17=0.8235$. Do not confuse the probability of white given B with the probability of B given white.

First update the source. The probability of the first white is $z=3/10\cdot2/5+(1-3/10)\cdot4/5=17/25$. After observing it, the weights of A and B are $3/17$ and $14/17$: divide each corresponding contribution by $z$. They sum to $1$.

When retaining the urn, use the updated weights and the within-urn probabilities of the second white: $3/17\cdot(1/4)+14/17\cdot(3/4)=45/68$. Without replacement, both the white count and the total decrease by one.

The answer $0.6618$ lies between the two within-urn probabilities for the second white: this checks the weighted average.

Convert proportions to counts: A and B have $0.84\cdot(100+150)=210$ on-time orders, giving B $210-90=120$. The total on time is $0.8\cdot500=400$, so C has $400-210=190$. Late counts by row are $10,30,60$. Check the row sums and the grand total.

Reduce the ratio $30:60$: answer 1:2.

Convert proportions to counts: A and B have $0.84\cdot(100+150)=210$ on-time orders, giving B $210-90=120$. The total on time is $0.8\cdot500=400$, so C has $400-210=190$. Late counts by row are $10,30,60$. Check the row sums and the grand total.

The condition leaves only $400$ on-time orders, of which $90$ came from A. The required proportion is $90/400=0.2250$. The denominator counts on-time orders, not all orders.

$u\cdot v=2\cdot3\cdot(0.5)$. Expanding the vector dotted with itself gives $\|3u+(-2)v\|^2=(3)^22^2+(-2)^23^2+2(3)(-2)(2)(3)(0.5)$. Take the nonnegative root: $6.0000$. The cross term may be negative, but the whole squared norm is nonnegative.

Choose $u=x^n$ so differentiation lowers the degree, and $dv=e^{-1x}dx$, giving $v=e^{-1x}/-1$. For $n\ge1$, the lower boundary term is zero: $B_n=1^n e^{-1}/-1-(n/-1)B_{n-1}$. At $n=0$ the lower term is not zero: $B_0=(e^{-1}-1)/-1$. Calculate up to the required index: $B_0=0.632121$, $B_1=0.264241$, $B_2=0.160603$. Sum only indices $0,2$: $0.7927$. Check that $B_n>0$, since the integrand is positive. Proof checklist: the choice of $u,dv$, both boundary terms, the range $n\ge1$, and a separate initial value.

$L(p)=p^{5}(1-p)^{7}$; $\ell(p)=5\ln p+7\ln(1-p)$. $\ell\prime(p)=5/p-7/(1-p)=0$, $\hat p=5/12=0.4167$.

$\ell\prime\prime(p)=-5/p^2-7/(1-p)^2<0$: the log-likelihood is strictly concave, so the stationary point is the unique maximum. It lies in $(0,1)$; at both ends $L$ tends to zero. The exponent of $(1-p)$ counts failures, not all trials.

First arrange the ordinary books in $3!$ ways. There are $4$ gaps, including both ends. Put at most one rare book in each: choose $4$ gaps and order the rare books in the chosen positions. This gives $3!\binom{4}{4}4!=144$. Every valid arrangement uniquely recovers this construction.

There are $(3+4)!$ equally likely permutations in total. Divide favorable by total: $P=0.0286$. The identities of the books are counted consistently in numerator and denominator.

The weight numerators sum to $3$, so $\E T=(3/3)\mu=\mu$ for every $\mu$: the estimator is unbiased. Independence is used in the second part: $\Var(T)=4\cdot((-1)^2+(1)^2+(3)^2)/3^2=4.8889$. Square the denominator too. Check that the result is nonnegative. The proof checklist is linearity, the sum of weights, and equality for every parameter value.

$\E X^2=\int_0^1 x^2(\theta+1)x^\theta dx=(\theta+1)/(\theta+3)$. Equate it to $0.7$ and multiply by the denominator to obtain $\hat\theta=3.6667>-1$. This uses the mean of powers, not the power of the sample mean; these are different. Substitution into the theoretical moment reproduces $0.7$.

Multiply the observation densities: $L(\theta)=(\theta+1)^{7}\exp(-14\theta)$. Its logarithm $\ell=7\ln(\theta+1)+(-14)\theta$ has derivative $7/(\theta+1)+(-14)$. Setting it to zero gives $\hat\theta=-0.5000$. The second derivative is $-7/(\theta+1)^2<0$, so the function is strictly concave on $\theta>-1$. The log-likelihood tends to $-\infty$ both near $-1$ and at infinity, establishing a global maximum.

Multiply the observation densities: $L(\theta)=(\theta+1)^{7}\exp(-14\theta)$. Its logarithm $\ell=7\ln(\theta+1)+(-14)\theta$ has derivative $7/(\theta+1)+(-14)$. Setting it to zero gives $\hat\theta=-0.5000$. The second derivative is $-7/(\theta+1)^2<0$, so the function is strictly concave on $\theta>-1$. The log-likelihood tends to $-\infty$ both near $-1$ and at infinity, establishing a global maximum.

Integrating the density gives $F(x)=x^{\theta+1}$ inside the support, $0$ for $x\le0$, and $1$ for $x\ge1$. A quantile satisfies $F(q)=0.25$, hence $q=0.25^{1/(\theta+1)}$. Substitute the unrounded parameter estimate: $\hat q=0.0625$. Check that $0<\hat q<1$ and the CDF at this point equals the required level.

Row $i$ of the outer product is $u_i v^T$: $uv^T=\begin{pmatrix}2&1&0\\-2&-1&0\\4&2&0\end{pmatrix}$. This is a matrix, whereas $v^Tu$ is a scalar. Add ones only on the diagonal: $A=\begin{pmatrix}3&1&0\\-2&0&0\\4&2&1\end{pmatrix}$.

Expand along the first row, retaining the alternating cofactor signs: $\det A=(3)(0)+(1)(2)+(0)(-4)=2$. Each second factor is the remaining minor determinant with sign $(-1)^{1+j}$. The nonzero determinant allows us to seek an inverse.

Row $i$ of the outer product is $u_i v^T$: $uv^T=\begin{pmatrix}2&1&0\\-2&-1&0\\4&2&0\end{pmatrix}$. This is a matrix, whereas $v^Tu$ is a scalar. Add ones only on the diagonal: $A=\begin{pmatrix}3&1&0\\-2&0&0\\4&2&1\end{pmatrix}$.

Write $[A\mid I]$. Apply these operations to every column of the augmented matrix, in order: $R_1\leftarrow R_1/(3)$; $R_2\leftarrow R_2-(-2)R_1$; $R_3\leftarrow R_3-(4)R_1$; $R_2\leftarrow R_2/(2/3)$; $R_1\leftarrow R_1-(1/3)R_2$; $R_3\leftarrow R_3-(2/3)R_2$. The left side becomes $I$ and the right side is $A^{-1}=\begin{pmatrix}0&-1/2&0\\1&3/2&0\\-2&-1&1\end{pmatrix}$. Add only the diagonal: $\tr(A^{-1})=5/2=2.5000$. Multiplication checks $AA^{-1}=I$ and can reveal an error in any column; checking only the determinant cannot do this.

Let $H(z)=\int_0^z e^{-t^2}dt$. The fundamental theorem gives $H\prime(z)=e^{-z^2}$. $F(x)=H(x^3)$, so apply the chain rule. Thus $F\prime(x)=e^{-(x^3)^2}3x^{2}$. At $x=-1$ the answer is $1.1036$. We need the local rate of area accumulation, not an explicit expression for the entire area.

Write $1/[x^2(x+2)]=A/x+B/x^2+C/(x+2)$. Multiplying by the denominator gives $1=A x(x+2)+B(x+2)+Cx^2$. Setting $x=0$ gives $B=1/2$; setting $x=-2$ gives $C=1/2^2$; the coefficient of $x^2$ gives $A=-1/2^2$. Substitution of all coefficients recovers the numerator $1$. An antiderivative is $H(x)=\ln((x+2)/x)/2^2-1/(2x)$. First work on a finite interval to get $H(T)-H(2)$. The logarithms are already combined: as $T\to\infty$, their ratio tends to $1$, so $H(T)\to0$. The original positive integrand is also bounded by $1/x^3$, confirming convergence. Answer: $0.0767$. Do not evaluate separately divergent logarithms as a difference of infinities.