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Under independence, there are two disjoint ways to score two goals in total. One game can contain both, with probabilityAlternatively, two different games can contain one goal each, with probabilityAdding givesThis also follows from the Sum of independent Poisson random variables: the season total is Poisson with parameter $10\lambda$, whose probability at two is $e^{-10\lambda }(10\lambda {)}^2/2!$.

The intended season calculation needs the additional assumption that scores in different games are independent random variables. The printed statement gives their common one-game distribution but does not explicitly state independence; identical marginals alone do not determine season probabilities.

Under independence, each game has probability $1-P(X=0)=1-e^{-\lambda }$ of at least one goal, soWithout independence this product is not determined by the supplied marginal distribution.

The slope of the particle's straight trajectory is $\tan {\theta }_i=y_i/x_0$, soThe endpoint directions $\pm \pi /2$ are parallel to the screen and do not give finite collisions, but each has zero probability in the continuous angular model. The relation therefore holds almost surely.

The angular interval has width $\pi$, so its continuous uniform distribution has probability density functionIntegrating this probability density function gives the cumulative distribution functionValues of the probability density function at the two individual endpoints may be assigned arbitrarily without changing any probability.

Multiply the production probability by the conditional stopping probability:

Distinguish the production type as well as the stopping position. The sample space has the six outcomesHere $x_k$ in an event means $X=x_k$. Their joint probabilities are $p{g}_k$ and $(1-p)d_k$, respectively. The conditional stopping laws obey ${\sum }_{k=0}^2{g}_k={\sum }_{k=0}^2{d}_k=1$; with the stated strict positivity all six outcomes are possible.

Divide the relevant joint probability by the marginal stopping probability. By Bayes' theorem,The denominator is positive under the stated assumptions.

The condition $X>x_0$ selects positions $x_1$ and $x_2$. Their combined good-ball joint probability is $p(g_1+g_2)$, and their total probability is $p(g_1+g_2)+(1-p)(d_1+d_2)$. Thus

There are $99$ possible stopping points, so the equal defective-ball probabilities are $d_k=1/99$. With $g_0>0$, the desired conditional probability givesThus, for a fixed good-ball stopping law,Since $g_0\le 1$, the smallest possible production probability over admissible stopping laws isIf strict positivity of every $g_k$ is additionally imposed in this general case, then $g_0<1$ and one only has the infimum $1/2$, approached as $g_0\to 1$. The paper states that strict-positivity assumption explicitly for part (a), not for the general case in part (b). Without choosing or bounding $g_0$, the exact fixed-law answer remains $1/(1+g_0)$.

The event includes stopping positions zero through $k$. Applying Bayes' theorem to their joint and marginal probabilities givesThis conditional probability is defined whenever its conditioning event has positive probability. At $k=n$, it reduces to $p$.

For $0\le k\le n$, sum the marginal stopping probabilities over the selected tail:Equivalently it is $p(1-{\sum }_{j=0}^k{g}_j)+(1-p)(1-{\sum }_{j=0}^k{d}_j)$. At $k=n$, both empty tail sums are zero, as required.

Use the intended model of uniform draws and a replacement decision independent of the first ball's colour, with probability $r$. The marginal specification $P(R)=r$ alone would not determine the experiment if replacement depended on that colour. This independence assumption is used throughout the question.

Write ${\overline{B}}_j$ for a non-blue draw and $\overline{R}$ for no replacement. The sample space consists of the eight ordered outcomes obtained by choosing one of $B_1,{\overline{B}}_1$, then one of $R,\overline{R}$, then one of $B_2,{\overline{B}}_2$. The chain rule for probabilities gives their probabilities:For example, the first entry is $(2/5)r(2/5)$. The fifth is $(2/5)(1-r)(1/4)$ because one blue ball has been removed. With no replacement after a non-blue first draw, the two remaining colours each have two balls, explaining the last two entries. The first four entries sum to $r$ and the last four to $1-r$, so the full sample space has total probability one. At $r=0$ or $1$, the corresponding impossible outcomes simply have zero probability.

Condition on whether replacement occurs. Both alternatives have blue second-draw probability $2/5$, so the law of total probability gives

For sampling without replacement, sum the two outcomes with a blue second draw:The marginal blue probability equals the initial blue fraction even though the two draws are dependent.

By the definition of conditional probability, or Bayes' theorem,Its denominator is positive throughout the specified range. Observing two blue balls raises the probability of replacement for $0<r<1$, because replacement makes a repeated colour more likely.

The event of two blue draws splits into disjoint replacement and no-replacement cases. Their probabilities from the sample space are $4r/25$ and $(1-r)/10$. Hence

Let $C={\overline{B}}_1\cap B_2$. Under replacement its joint probability iswhereas without replacement it isIn the latter case the first removed ball is non-blue, so all $N_B$ blue balls remain. Applying Bayes' theorem and cancelling the positive common factors yields the posterior replacement probability after different-colour draws:It does not depend on $N_B$. To prove the requested bound directly,Equality holds at $r=0$ and $r=1$, and the inequality is strict between them. Its sketch joins $(0,0)$ to $(1,1)$, stays below the diagonal, and is increasing and convex becauseThe interpretation is that removing a non-blue ball increases the chance of blue on the second draw; seeing the two different colours therefore favours the no-replacement explanation.

Each position has apple probability $1/2$. Applying Bayes' theorem to the preceding conditional probability givesThis also agrees with exchangeability of the second and third positions.

Random sampling without replacement assigns each of the ten distinguishable pieces to each lunch with the same probability. Five pieces of each type therefore giveOne can check the second probability by conditioning on the first lunch:The identical marginal probabilities do not imply independence of the draws.

By the inclusion-exclusion principle for probabilities,After an initial banana, five of the nine remaining pieces are apples, so the intersection probability is $(1/2)(5/9)=5/18$. ConsequentlyAs a check, the complementary event is an apple followed by a banana, which also has probability $5/18$.