Binomial Probability and Bernoulli Trials: Difficult Problems with Solutions

Solution:
Let [tex]p[/tex] be the probability of a hit with one shot; then a miss has probability [tex]1-p[/tex]. The shots are independent, so the probability of no hit in [tex]4[/tex] shots is [tex](1-p)^4[/tex]. The event "at least one hit" is the complement of "no hit", hence
[tex]1-(1-p)^4=0.9984\ \Rightarrow\ (1-p)^4=0.0016=0.2^4[/tex]
Since [tex]1-p\ge 0[/tex], we take the non-negative fourth root:
[tex]1-p=0.2\ \Rightarrow\ p=0.8[/tex]
The probability of a hit with one shot is [tex]0.8[/tex].

Solution:
The observations are Bernoulli trials with [tex]p=0.8[/tex] and [tex]q=1-p=0.2[/tex]. The probability of no detection in [tex]n[/tex] observations is [tex]0.2^n[/tex], so the probability of at least one detection is [tex]1-0.2^n[/tex]. We need
[tex]1-0.2^n\ge 0.99\ \Leftrightarrow\ 0.2^n\le 0.01[/tex]
The power [tex]0.2^n[/tex] decreases as [tex]n[/tex] grows. We check the first values:
[tex]0.2^2=0.04\gt 0.01 , \qquad 0.2^3=0.008\le 0.01[/tex]
So [tex]2[/tex] observations are not enough, while [tex]3[/tex] observations give probability [tex]1-0.008=0.992\ge 0.99[/tex]. The least number of observations is [tex]3[/tex].

Solution:
We have [tex]n=5[/tex] Bernoulli trials with success probability [tex]p=0.8[/tex] and [tex]q=0.2[/tex]. The probability of exactly [tex]k[/tex] cured patients is
[tex]P_5(k)=\binom{5}{k}\times 0.8^k\times 0.2^{5-k}[/tex]
Compare consecutive probabilities:
[tex]\frac{P_5(k+1)}{P_5(k)}=\frac{5-k}{k+1}\times \frac{0.8}{0.2}=\frac{4(5-k)}{k+1}[/tex]
This ratio is greater than [tex]1[/tex] exactly when [tex]20-4k\gt k+1[/tex], i.e. [tex]k\lt 3.8[/tex]. Hence the probabilities increase up to [tex]k=4[/tex] and decrease after it:
[tex]P_5(0)\lt P_5(1)\lt \ldots\lt P_5(4)\gt P_5(5)[/tex]
Check:
[tex]P_5(3)=10\times 0.8^3\times 0.2^2=0.2048 , \quad P_5(4)=5\times 0.8^4\times 0.2=0.4096 , \quad P_5(5)=0.8^5=0.32768[/tex]
The most probable number of cured patients is [tex]4[/tex].

Solution:
Since each ball is returned, the [tex]8[/tex] draws are independent, and in each of them a black ball appears with the same probability:
[tex]p=\frac{5}{20}=\frac{1}{4} , \quad q=1-p=\frac{3}{4}[/tex]
So these are Bernoulli trials, and the probability of exactly [tex]k[/tex] black balls is [tex]P_8(k)=\binom{8}{k}p^kq^{8-k}[/tex]. Compare consecutive probabilities:
[tex]\frac{P_8(k+1)}{P_8(k)}=\frac{8-k}{k+1}\times \frac{p}{q}=\frac{8-k}{3(k+1)}[/tex]
This ratio is greater than [tex]1[/tex] exactly when [tex]8-k\gt 3k+3[/tex], i.e. [tex]k\lt \frac{5}{4}[/tex], so only for [tex]k=0[/tex] and [tex]k=1[/tex]; for [tex]k\ge 2[/tex] it is less than [tex]1[/tex]. Hence
[tex]P_8(0)\lt P_8(1)\lt P_8(2)\gt P_8(3)\gt \ldots\gt P_8(8)[/tex]
The most probable number of black balls is [tex]2[/tex]; its probability is [tex]P_8(2)=28\times \frac{1}{16}\times \frac{729}{4096}\approx 0.311[/tex].

Solution:
Let [tex]u[/tex] be the number of days with a rise and [tex]d[/tex] the number of days with a fall. Then
[tex]u+d=10 , \quad u-d=-4\ \Rightarrow\ u=3 , \quad d=7[/tex]
The [tex]10[/tex] days are independent repetitions of the same trial with probability [tex]p=\frac{1}{2}[/tex] of a rise, so by Bernoulli's formula the probability of exactly [tex]3[/tex] rises is
[tex]P=\binom{10}{3}\left(\frac{1}{2}\right)^{3}\left(\frac{1}{2}\right)^{7}=\frac{120}{1024}=\frac{15}{128}[/tex]
The probability that the price has fallen by [tex]4[/tex] points is [tex]\frac{15}{128}[/tex].

Solution:
In one kick there are [tex]3\times 3=9[/tex] equally likely pairs of directions (Ivancho's and Mariyka's), and in [tex]3[/tex] of them the directions coincide. So a kick is saved with probability [tex]\frac{3}{9}=\frac{1}{3}[/tex] and is a goal with probability
[tex]p=1-\frac{1}{3}=\frac{2}{3}[/tex]
The [tex]5[/tex] kicks are independent repetitions of the same trial (Bernoulli trials with [tex]n=5[/tex] and [tex]p=\frac{2}{3}[/tex]), so the probability of exactly [tex]k[/tex] goals is
[tex]P_5(k)=\binom{5}{k}\left(\frac{2}{3}\right)^k\left(\frac{1}{3}\right)^{5-k}=\binom{5}{k}\frac{2^k}{3^5}[/tex]
Ivancho wins with [tex]3[/tex], [tex]4[/tex] or [tex]5[/tex] goals:
[tex]P=\binom{5}{3}\frac{2^3}{3^5}+\binom{5}{4}\frac{2^4}{3^5}+\binom{5}{5}\frac{2^5}{3^5}=\frac{80+80+32}{243}=\frac{192}{243}=\frac{64}{81}[/tex]
The probability that Ivancho wins is [tex]\frac{64}{81}[/tex].

Solution:
In one draw all [tex]\binom{80}{20}[/tex] sets of [tex]20[/tex] numbers are equally likely, and [tex]\binom{79}{19}[/tex] of them contain the chosen number. So in one draw the chosen number is drawn with probability
[tex]p=\frac{\binom{79}{19}}{\binom{80}{20}}=\frac{20}{80}=\frac{1}{4}[/tex]
The draws are independent, so in [tex]n[/tex] draws the number is never drawn with probability [tex]\left(\frac{3}{4}\right)^n[/tex] and is drawn at least once with probability [tex]1-\left(\frac{3}{4}\right)^n[/tex]. We need
[tex]1-\left(\frac{3}{4}\right)^n\gt 0.9\ \Leftrightarrow\ \left(\frac{3}{4}\right)^n\lt 0.1[/tex]
The power [tex]\left(\frac{3}{4}\right)^n[/tex] decreases as [tex]n[/tex] grows. For [tex]n=8[/tex] the condition fails (the probability is about [tex]0.8999[/tex]):
[tex]\left(\frac{3}{4}\right)^8=\frac{6561}{65536}\gt\frac{6553.6}{65536}=0.1[/tex]
For [tex]n=9[/tex] it holds (the probability is about [tex]0.925\gt 0.9[/tex]):
[tex]\left(\frac{3}{4}\right)^9=\frac{19683}{262144}\lt\frac{26214.4}{262144}=0.1[/tex]
The least number of draws is [tex]9[/tex].

Solution:
Let [tex]n[/tex] be the number of throws and [tex]A[/tex] the event "a six comes up at least once". The throws are independent, and in each of them a six does not come up with probability [tex]\frac{5}{6}[/tex]. The complementary event "no six in [tex]n[/tex] throws" has probability [tex]\left(\frac{5}{6}\right)^n[/tex], so
[tex]P(A)=1-\left(\frac{5}{6}\right)^n\ge 0.95\iff\left(\frac{5}{6}\right)^n\le 0.05[/tex]
Take decimal logarithms of both sides (the function [tex]\lg[/tex] is increasing):
[tex]n\lg\frac{5}{6}\le\lg 0.05[/tex]
With the given values
[tex]\lg\frac{5}{6}=\lg\frac{10}{12}=1-2\lg 2-\lg 3\approx 1-0.6020-0.4771=-0.0791[/tex]
[tex]\lg 0.05=\lg\frac{10}{200}=1-\lg 2-2\approx -1.3010[/tex]
Since [tex]\lg\frac{5}{6}\lt 0[/tex], dividing by it reverses the inequality:
[tex]n\ge\frac{\lg 0.05}{\lg\frac{5}{6}}\approx\frac{-1.3010}{-0.0791}\approx 16.4[/tex]
The least such integer is [tex]n=17[/tex]. Check: [tex]\left(\frac{5}{6}\right)^{16}\approx 0.054\gt 0.05[/tex] and [tex]\left(\frac{5}{6}\right)^{17}\approx 0.045\lt 0.05[/tex]. So the die must be thrown at least [tex]17[/tex] times.

Solution:
Fix the chosen page. For each of the six figures, "the figure is on this page" is a trial with probability of success [tex]p=\frac{1}{120}[/tex] and of failure [tex]q=\frac{119}{120}[/tex], and the six trials are independent. So the number of figures on the page is the number of successes in [tex]6[/tex] Bernoulli trials, and
[tex]P=\sum_{k=3}^{6}\binom{6}{k}p^kq^{6-k}=\binom{6}{3}p^3q^3+\binom{6}{4}p^4q^2+\binom{6}{5}p^5q+\binom{6}{6}p^6[/tex]
Here [tex]p^kq^{6-k}=\frac{119^{6-k}}{120^6}[/tex] and [tex]\binom{6}{3}=20[/tex], [tex]\binom{6}{4}=15[/tex], [tex]\binom{6}{5}=6[/tex], [tex]\binom{6}{6}=1[/tex], so
[tex]P=\frac{20\times 119^{3}+15\times 119^{2}+6\times 119+1}{120^{6}}\approx 1.14\times 10^{-5}[/tex]