Answers Probability | Stats Book

We need to use $P={t\choose n}\times(c)^n\times{(1-c)^{t-n}$ , t = 10, n = 3 and c = 1/10. Therefore: $P={10\choose 3}\times(\frac{1}{10})^3\times{(\frac{9}{10})^{10-3}=\frac{10!}{7!\times3!}\times\frac{1^3}{10^3}\times{\frac{9^7}{10^7}=\frac{10!}{7!\times3!}\times\frac{9^7}{10^{10}}$ $=\frac{10\times9\times8}{3\times2\times1}\times\frac{9^7}{10^{10}}=\frac{3443737680}{60000000000}\app0.0574\app5.7%$
The total number of combinations is: We start by painting 9 of the 20 objects red. There are ${20 \choose 9}$ possible combinations of doing that. Of the remaining 11 objects, we paint 8 white. That can be done in ${11 \choose 8}$ possible combinations. The final three objects are painted blue. Obviously there is only one possible combination of doing that. The total number of possible combinations is therefore: ${20 \choose 9}\times{{20-9}\choose8}\times{{20-9-8}\choose3}={20 \choose 9}\times{{11}\choose8}\times{{3}\choose3}$ $=\frac{20!}{11!\times9!}\times\frac{11!}{3!\times8!}\times\frac{3!}{3!}=\frac{20!}{9!\times3!\times8!}=27713400$ . One might think that the answer would have been different if we would have started by painting the 3 blue objects. The following formula describes the possible combinations by first painting 3 objects blue, followed by 8 objects white: ${20 \choose 3}\times{{20-3}\choose8}\times{{20-3-8}\choose9}={20 \choose 3}\times{{17}\choose8}\times{{9}\choose9}$ $=\frac{20!}{17!\times3!}\times\frac{17!}{9!\times8!}\times\frac{9!}{9!}=\frac{20!}{3!\times8!\times9!}=27713400$ . Indeed, if we would have started by painting the 8 white objects, followed by nine objects red, the formula would have been ${20 \choose 8}\times{{20-8}\choose9}\times{{20-8-9}\choose3}={20 \choose 8}\times{{12}\choose9}\times{{3}\choose3}$ $=\frac{20!}{12!\times8!}\times\frac{12!}{3!\times9!}\times\frac{3!}{3!}=\frac{20!}{8!\times9!\times3!}=27713400$ . So, it doesn’t make any difference where you start. We have shown that: ${20 \choose 9}\times{{11}\choose8}\times{{3}\choose3}={20 \choose 3}\times{{17}\choose8}\times{{9}\choose9}={20 \choose 8}\times{{12}\choose9}\times{{3}\choose3}$
This is very similar to the example given in the main text. The probability is: $P=\frac{{4\choose 1 }\times{1\choose 1}\times{3 \choose 2}\times{44\choose1}}{52\choose 5}$ $=\frac{\frac{4!}{3!\times1!}\times1\times\frac{3!}{1!\times2!}\times\frac{44!}{43!\times1!}}{\frac{52!}{47!\times5!}}=\frac{4!\times3!\times44!\times47!\times5!}{3!\times2!\times43!\times52!}$ $=\frac{47!\times44!\times5!\times4!\times3!}{52!\times43!\times3!\times2!}=\frac{44\times5\times4\times3\times4\times3\times2}{52\times51\times50\times49\times48}=\frac{63360}{311875200}\app0.000203$
$P=\frac{{4\choose 1 }\times{1\choose 1}\times{3 \choose 0}\times{44\choose3}}{52\choose 5}$ $=\frac{\frac{4!}{3!\times1!}\times1\times1\times\frac{44!}{41!\times3!}}{\frac{52!}{47!\times5!}}=\frac{4!\times44!\times47!\times5!}{3!\times1!\times41!\times3!\times52!}$ $= \frac{44\times43\times42\times5\times4\times4}{52\times51\times50\times49\times48}=\frac{6357120}{311875200}\app0.0204\app2%$
This probability can be calculated by adding the probability of 1 king plus the probability on two kings plus the probability on three kings plus the probability of four kings: $P=\frac{{4\choose 1 }\times{48\choose 4}}{52\choose 5}+\frac{{4\choose 2 }\times{48\choose 3}}{52\choose 5}+\frac{{4\choose 3 }\times{48\choose 2}}{52\choose 5}+$ $\frac{{4\choose 4 }\times{48\choose 1}}{52\choose 5}$ $=\frac{\frac{4!}{3!\times1!}\times\frac{48!}{44!\times4!}}{\frac{52!}{47!\times5!}}+\frac{\frac{4!}{2!\times2!}\times\frac{48!}{45!\times3!}}{\frac{52!}{47!\times5!}}$ $+\frac{\frac{4!}{1!\times3!}\times\frac{48!}{46!\times2!}}{\frac{52!}{47!\times5!}}+\frac{1\times\frac{48!}{47!\times1!}}{\frac{52!}{47!\times5!}}$ $=\frac{\frac{48\times47\times46\times45}{3\times2}+\frac{48\times47\times46}{1}+\frac{48\times47\times2}{1}+\frac{48}{1}}{\frac{52\times51\times50\times49\times48}{5\times4\times3\times2}}$ $=\frac{886656}{2598960}\app0.341158$ . It would have been simpler to calculate the probability that no king is drawn and subtracting this from 1: $P=1-\frac{{4\choose 0 }\times{48\choose 5}}{52\choose 5}=1-(\frac{48!}{43!\times5!}\times\frac{47!\time5!}{52!})$ $=1-(\frac{48!\times47!\times5!}{52!\times43!\times5!})=1-(\frac{48!\times47!}{52!\times43!})$ $=1-(\frac{47\times46\times45\times44}{52\times51\times50\times49})=1-\frac{4280760}{6497400}\app1-0.659\app0.341158$
$P=\frac{{6\choose 6 }\times{43\choose 0}}{49\choose 6}=\frac{1}{\frac{49!}{43!\times6!}}=\frac{1}{\frac{49\times48\times47\times46\times45\times44}{6\times5\times4\times3\times2}}\app0.0000000715$