Binomial Probability and Bernoulli Trials: Problems with Solutions

Solution:
Since the ball is put back after every draw, the [tex]10[/tex] draws are independent repetitions of the same trial. In one draw the red ball is drawn with probability [tex]p=\frac{1}{3}[/tex], so it is not drawn with probability
[tex]q=1-\frac{1}{3}=\frac{2}{3}[/tex]
The red ball is never drawn exactly when this happens in all [tex]10[/tex] draws (zero successes in a Bernoulli scheme with [tex]n=10[/tex], [tex]p=\frac{1}{3}[/tex]), so
[tex]P=q^{10}=\left(\frac{2}{3}\right)^{10}=\frac{1024}{59049}\approx 0.0173[/tex]

Solution:
Each child is an independent trial with success (blond) probability [tex]p=\frac{1}{3}[/tex] and failure probability [tex]q=\frac{2}{3}[/tex]. This is a Bernoulli scheme with [tex]n=4[/tex] trials, and we need exactly [tex]k=2[/tex] successes. The two blond children can be chosen in [tex]\binom{4}{2}=6[/tex] ways, and each such choice has probability [tex]p^2q^2[/tex]. Hence
[tex]P_4(2)=\binom{4}{2}p^2q^2=6\times \frac{1}{9}\times \frac{4}{9}=\frac{24}{81}=\frac{8}{27}[/tex]

Solution:
The games are independent trials in which the first player wins with probability [tex]p=\frac{1}{2}[/tex], so the Bernoulli formula applies:
[tex]P_n(k)=\binom{n}{k}p^k(1-p)^{n-k}=\frac{\binom{n}{k}}{2^n}[/tex]
Hence
[tex]P_4(3)=\frac{\binom{4}{3}}{2^4}=\frac{4}{16}=\frac{1}{4},\qquad P_8(5)=\frac{\binom{8}{5}}{2^8}=\frac{56}{256}=\frac{7}{32}[/tex]
Since [tex]\frac{1}{4}=\frac{8}{32}\gt\frac{7}{32}[/tex], winning exactly [tex]3[/tex] of [tex]4[/tex] games is more probable. Note that [tex]\frac{3}{4}[/tex] and [tex]\frac{5}{8}[/tex] are only the shares of games won, not probabilities.

Solution:
Checking the [tex]6[/tex] items is a sequence of [tex]n=6[/tex] independent trials, each a success (good item) with probability [tex]p=0.9[/tex] and a failure with probability [tex]q=0.1[/tex]. The [tex]4[/tex] good items can occupy [tex]\binom{6}{4}=15[/tex] sets of positions, and each such arrangement has probability [tex]p^4q^2[/tex]. By the Bernoulli formula
[tex]P_6(4)=\binom{6}{4}p^4q^2=15\times 0.6561\times 0.01=0.098415[/tex]

Solution:
The [tex]5[/tex] throws are independent repetitions of the same trial with success probability [tex]p=0.9[/tex] and failure probability [tex]q=0.1[/tex], so this is a Bernoulli scheme and [tex]P_5(k)=\binom{5}{k}p^kq^{5-k}[/tex]. At least [tex]4[/tex] successes means exactly [tex]4[/tex] or exactly [tex]5[/tex] successes:
[tex]P_5(4)=\binom{5}{4}\times 0.9^4\times 0.1=5\times 0.6561\times 0.1=0.32805[/tex]
[tex]P_5(5)=0.9^5=0.59049[/tex]
These two events are mutually exclusive, so
[tex]P=P_5(4)+P_5(5)=0.32805+0.59049=0.91854[/tex]

Solution:
The [tex]11[/tex] tosses are independent repetitions of the same trial, and in each toss heads comes up with probability [tex]p=\frac{1}{2}[/tex] and does not with probability [tex]q=\frac{1}{2}[/tex]. In this Bernoulli scheme with [tex]n=11[/tex] the probability of exactly [tex]k[/tex] heads is
[tex]P_{11}(k)=\binom{11}{k}p^kq^{11-k}=\frac{\binom{11}{k}}{2^{11}}[/tex]
At most [tex]4[/tex] times means [tex]k\in\left\{0, 1, 2, 3, 4\right\}[/tex], and these cases are mutually exclusive:
[tex]P=\frac{\binom{11}{0}+\binom{11}{1}+\binom{11}{2}+\binom{11}{3}+\binom{11}{4}}{2^{11}}=\frac{1+11+55+165+330}{2048}=\frac{562}{2048}=\frac{281}{1024}[/tex]

Solution:
Each coin shows heads with probability [tex]\frac{1}{2}[/tex], independently of the others, so this is a Bernoulli scheme with [tex]n=100[/tex] and [tex]p=q=\frac{1}{2}[/tex]. By the Bernoulli formula
[tex]P_{100}(50)=\binom{100}{50}\left(\frac{1}{2}\right)^{50}\left(\frac{1}{2}\right)^{50}=\frac{\binom{100}{50}}{2^{100}}[/tex]
Equivalently, there are [tex]2^{100}[/tex] equally likely outcomes, and the [tex]50[/tex] coins showing heads can be chosen in [tex]\binom{100}{50}[/tex] ways. This probability is only about [tex]0.08[/tex], far from [tex]\frac{1}{2}[/tex]: although [tex]50[/tex] is the most probable number of heads, each single value of the number of heads is unlikely.

Solution:
The answers to the [tex]100[/tex] questions are [tex]100[/tex] independent repetitions of the same trial: success (correct answer) with probability [tex]p=\frac{1}{4}[/tex] and failure with probability [tex]q=\frac{3}{4}[/tex]. By the Bernoulli formula, the probability of exactly [tex]k=90[/tex] successes in [tex]n=100[/tex] trials is
[tex]P_{100}(90)=\binom{100}{90}p^{90}q^{10}=\binom{100}{90}\times \frac{1}{4^{90}}\times \frac{3^{10}}{4^{10}}=\binom{100}{90}\times \frac{3^{10}}{4^{100}}[/tex]
The factor [tex]\binom{100}{90}[/tex] counts the ways to choose which [tex]90[/tex] questions are answered correctly, and the factor [tex]q^{10}[/tex] is needed because the other [tex]10[/tex] answers must be wrong.