Binomial Probability and Bernoulli Trials: Very Difficult Problems with Solutions

Solution:
Each of the four chosen customers is, independently of the others, a woman, a man or a child with probabilities [tex]0.6[/tex], [tex]0.3[/tex] and [tex]0.1[/tex], respectively. Let [tex]M[/tex] be the event "there is no man among the four" and [tex]C[/tex] the event "there is no child among the four". The event [tex]M\cap C[/tex] means that all four are women. Then
[tex]P(M)=0.7^4=0.2401, \quad P(C)=0.9^4=0.6561, \quad P(M\cap C)=0.6^4=0.1296[/tex]
The required event is the complement of [tex]M\cup C[/tex], so by the addition rule
[tex]P=1-\left(P(M)+P(C)-P(M\cap C)\right)=1-0.2401-0.6561+0.1296=0.2334[/tex]

Solution:
The checking stops before the sixth part exactly when the fourth good part appears among the first five, i.e. when the first five parts contain at least [tex]4[/tex] good ones. Hence all the parts are checked exactly when the first five parts contain at most [tex]3[/tex] good ones, i.e. at least [tex]2[/tex] defective ones (if the fourth good part is the sixth one, it is checked as well).
The first five parts form [tex]5[/tex] independent Bernoulli trials with probability [tex]0.1[/tex] of a defect. Let [tex]P_5(k)[/tex] be the probability of exactly [tex]k[/tex] defective parts among them. We pass to the complementary event (at most one defective part):
[tex]P_5(0)=0.9^5=0.59049, \quad P_5(1)=\binom{5}{1}\times 0.1\times 0.9^4=0.32805[/tex]
Therefore
[tex]P=1-P_5(0)-P_5(1)=1-0.59049-0.32805=0.08146[/tex]

Solution:
Let [tex]q=1-p[/tex] and let [tex]P_{39}(k)=\binom{39}{k}p^kq^{39-k}[/tex] be the probability of exactly [tex]k[/tex] successes. Then
[tex]\frac{P_{39}(k+1)}{P_{39}(k)}=\frac{39-k}{k+1}\times \frac{p}{q}[/tex]
This ratio decreases as [tex]k[/tex] grows, so the probabilities first increase and then decrease. Hence the consecutive numbers [tex]25[/tex] and [tex]26[/tex] are both most probable exactly when [tex]P_{39}(26)=P_{39}(25)[/tex], i.e. when the ratio for [tex]k=25[/tex] equals [tex]1[/tex]:
[tex]\frac{39-25}{25+1}\times \frac{p}{1-p}=1\iff 14p=26(1-p)\iff 40p=26\iff p=\frac{13}{20}[/tex]
This agrees with the general rule: the most probable number [tex]k_0[/tex] of successes in [tex]n[/tex] trials satisfies [tex](n+1)p-1\le k_0\le (n+1)p[/tex], and there are two such numbers exactly when [tex](n+1)p[/tex] is an integer; here [tex]n=39[/tex] and [tex](n+1)p=40\times \frac{13}{20}=26[/tex].

Solution:
Suppose the first player wins the match and consider the game in which he gets his [tex]m[/tex]-th win. At that moment the second player has [tex]k[/tex] wins, where [tex]k\in\{0, 1, \ldots, n-1\}[/tex] (with [tex]n[/tex] wins he would have already won the match). This happens exactly when game number [tex]m+k[/tex] is won by the first player and among the previous [tex]m+k-1[/tex] games exactly [tex]k[/tex] are won by the second player. Since the games are independent, by the Bernoulli formula the probability of this is
[tex]\binom{m+k-1}{k}p^{m-1}q^{k}\times p=\binom{m+k-1}{k}p^{m}q^{k}[/tex]
These events for different [tex]k[/tex] are mutually exclusive, so
[tex]P=\sum_{k=0}^{n-1}\binom{m+k-1}{k}p^{m}q^{k}[/tex]
Check: the same number is the probability that the first player wins at least [tex]m[/tex] of [tex]m+n-1[/tex] games, because playing extra fictitious games after the match is decided does not change the winner. For example, for [tex]m=2[/tex], [tex]n=3[/tex], [tex]p=q=\frac{1}{2}[/tex] both ways give
[tex]\frac{1}{4}+\frac{2}{8}+\frac{3}{16}=\frac{11}{16}=\frac{\binom{4}{2}+\binom{4}{3}+\binom{4}{4}}{16}[/tex]

Solution:
Put [tex]A=(0, 0, 0)[/tex] and [tex]C=(1, 1, 1)[/tex], and call the number of coordinates of a vertex equal to [tex]1[/tex] its level. The levels are [tex]0[/tex], [tex]1[/tex], [tex]2[/tex], [tex]3[/tex], and [tex]C[/tex] is the only vertex of level [tex]3[/tex]. Each move along an edge changes the level by [tex]1[/tex]: from level [tex]0[/tex] the ant always goes to level [tex]1[/tex]; from level [tex]1[/tex] it goes to level [tex]0[/tex] with probability [tex]\frac{1}{3}[/tex] and to level [tex]2[/tex] with probability [tex]\frac{2}{3}[/tex]; from level [tex]2[/tex] it goes to [tex]C[/tex] with probability [tex]\frac{1}{3}[/tex] and back to level [tex]1[/tex] with probability [tex]\frac{2}{3}[/tex].
After an odd number of minutes the level is odd and after an even number it is even, so [tex]C[/tex] can be reached only at the end of an odd minute. After the first minute the ant is at level [tex]1[/tex]. Split the following time into blocks of two minutes. If the ant starts a block at level [tex]1[/tex], it reaches [tex]C[/tex] at the end of the block with probability
[tex]\frac{2}{3}\times \frac{1}{3}=\frac{2}{9}[/tex]
and otherwise (with probability [tex]\frac{1}{3}+\frac{2}{3}\times \frac{2}{3}=\frac{7}{9}[/tex]) it is again at level [tex]1[/tex]. Since the choices are independent, the blocks are independent repetitions of the same trial. Hence the probability that the ant has not passed through [tex]C[/tex] after [tex]2k+1[/tex] minutes (and also after [tex]2k+2[/tex] minutes) is [tex]\left(\frac{7}{9}\right)^k[/tex]. We need
[tex]1-\left(\frac{7}{9}\right)^k\gt\frac{9}{10}\iff 10\times 7^k\lt 9^k[/tex]
The ratio [tex]\frac{9^k}{7^k}[/tex] increases with [tex]k[/tex], and for [tex]k=9[/tex] and [tex]k=10[/tex] we have
[tex]10\times 7^9=403\,536\,070\gt 9^9=387\,420\,489[/tex]
[tex]10\times 7^{10}=2\,824\,752\,490\lt 9^{10}=3\,486\,784\,401[/tex]
So the least suitable [tex]k[/tex] is [tex]10[/tex], and the required number of minutes is [tex]2\times 10+1=21[/tex] (after [tex]20[/tex] minutes the probability is still [tex]1-\left(\frac{7}{9}\right)^9\lt 0.9[/tex]).