- We need to use:
\(Probability = {t \choose n} \cdot c^n \cdot{(1-c)^{t-n}} \)
t = 10, n = 3 and c = 1/10. Therefore
\(Probability = {10 \choose 3} \cdot {\frac{1}{10}}^3 \cdot{(1-{\frac{1}{10}})^{10-3}} \)
\(Probability = {10 \choose 3} \cdot {\frac{1}{10}}^3 \cdot{{\frac{9}{10}}^{7}} \)
\(Probability = \frac{10!}{7!\cdot3!}\cdot\frac{1}{10}^3\cdot{\frac{9}{10}^7} \)
\(Probability = \frac{10\cdot9\cdot8}{3\cdot2\cdot1}\cdot{\frac{9^7}{10^{10}}} \)
\(Probability = \frac{3443737680}{60000000000} \) ~ 0.0574 ~ 5.7%
2. The total number of combinations is: We start by painting 9 of the 20 injects 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:
\( {20 \choose 9}\cdot{{20-9}\choose8}\cdot{{20-9-8}\choose3}={20 \choose 9}\cdot{{11}\choose8}\cdot{{3}\choose3} \)
= \( \frac{20!}{11!\cdot9!}\cdot\frac{11!}{3!\cdot8!}\cdot\frac{3!}{3!}=\frac{20!}{9!\cdot3!\cdot8!} \) =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
\( {20 \choose 3}\cdot{{20-3}\choose8}\cdot{{20-3-8}\choose9}={20 \choose 3}\cdot{{17}\choose8}\cdot{{9}\choose9} \)
\( =\frac{20!}{17!\cdot3!}\cdot\frac{17!}{9!\cdot8!}\cdot\frac{9!}{9!}=\frac{20!}{3!\cdot8!\cdot9!} \)=27713400
Indeed, if we would have started by painting the 8 white objects, followed by nine objects red, the formula would have been:
\( {20 \choose 8}\cdot{{20-8}\choose9}\cdot{{20-8-9}\choose3}={20 \choose 8}\cdot{{12}\choose9}\cdot{{3}\choose3} \)
\( =\frac{20!}{12!\cdot8!}\cdot\frac{12!}{3!\cdot9!}\cdot\frac{3!}{3!}=\frac{20!}{8!\cdot9!\cdot3!} \)=27713400
So, it doesn’t make any difference where you start. We have shown that:
\( {20 \choose 9}\cdot{{11}\choose8}\cdot{{3}\choose3}={20 \choose 3}\cdot{{17}\choose8}\cdot{{9}\choose9}={20 \choose 8}\cdot{{12}\choose9}\cdot{{3}\choose3}\)
3. This is very similar to the example given in the main text. The probability is:
\( Probability=\frac{{4\choose 1 }\cdot{1\choose 1}\cdot{3 \choose 2}\cdot{44\choose1}}{52\choose 5} \)
\( Probability=\frac{\frac{4!}{3!\cdot1!}\cdot1\cdot\frac{3!}{1!\cdot2!}\cdot\frac{44!}{43!\cdot1!}}{\frac{52!}{47!\cdot5!}}=\frac{4!\cdot3!\cdot44!\cdot47!\cdot5!}{3!\cdot2!\cdot43!\cdot52!} \)
\(Probability=\frac{47!\cdot44!\cdot5!\cdot4!\cdot3!}{52!\cdot43!\cdot3!\cdot2!}=\frac{44\cdot5\cdot4\cdot3\cdot4\cdot3\cdot2}{52\cdot51\cdot50\cdot49\cdot48}=\frac{63360}{311875200} \)
\( Probability \) ~ 0.000203
4.
\( Probability=\frac{{4\choose 1 }\cdot{1\choose 1}\cdot{3 \choose 0}\cdot{44\choose3}}{52\choose 5} \)
\(Probability =\frac{\frac{4!}{3!\cdot1!}\cdot1\cdot1\cdot\frac{44!}{41!\cdot3!}}{\frac{52!}{47!\cdot5!}}=\frac{4!\cdot44!\cdot47!\cdot5!}{3!\cdot1!\cdot41!\cdot3!\cdot52!} \)
\(Probability = \frac{44\cdot43\cdot42\cdot5\cdot4\cdot4}{52\cdot51\cdot50\cdot49\cdot48}=\frac{6357120}{311875200} \) ~ 0.0204 ~ 2%
5. 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:
\( Probability=\frac{{4\choose 1 }\cdot{48\choose 4}}{52\choose 5}+\frac{{4\choose 2 }\cdot{48\choose 3}}{52\choose 5}+\frac{{4\choose 3 }\cdot{48\choose 2}}{52\choose 5}+ \frac{{4\choose 4 }\cdot{48\choose 1}}{52\choose 5} \)
\(Probability =\frac{\frac{4!}{3!\cdot1!}\cdot\frac{48!}{44!\cdot4!}}{\frac{52!}{47!\cdot5!}}+\frac{\frac{4!}{2!\cdot2!}\cdot\frac{48!}{45!\cdot3!}}{\frac{52!}{47!\cdot5!}} +\frac{\frac{4!}{1!\cdot3!}\cdot\frac{48!}{46!\cdot2!}}{\frac{52!}{47!\cdot5!}}+\frac{1\cdot\frac{48!}{47!\cdot1!}}{\frac{52!}{47!\cdot5!}} \)
\( Probability=\frac{\frac{48\cdot47\cdot46\cdot45}{3\cdot2}+\frac{48\cdot47\cdot46}{1}+\frac{48\cdot47\cdot2}{1}+\frac{48}{1}}{\frac{52\cdot51\cdot50\cdot49\cdot48}{5\cdot4\cdot3\cdot2}} \)
\(Probability =\frac{886656}{2598960} \) ~ 0.341158
It would have been easier to subtract the probability of no kings from one:
\( Probability=1-\frac{{4\choose 0 }\cdot{48\choose 5}}{52\choose 5}=1-(\frac{48!}{43!\cdot5!}\cdot\frac{47!\cdot5!}{52!}) \)
\(Probability =1-(\frac{48!\cdot47!\cdot5!}{52!\cdot43!\cdot5!})=1-(\frac{48!\cdot47!}{52!\cdot43!}) \)
\( Probabiliyt =1-(\frac{47\cdot46\cdot45\cdot44}{52\cdot51\cdot50\cdot49}) \)
\( Probability=1-\frac{4280760}{6497400} \) ~ 1-0.659 ~ 0.341158
6.
\(Probability =\frac{{6\choose 6 }\cdot{43\choose 0}}{49\choose 6}=\frac{1}{\frac{49!}{43!\cdot6!}}=\frac{1}{\frac{49\cdot48\cdot47\cdot46\cdot45\cdot44}{6\cdot5\cdot4\cdot3\cdot2}} \) ~ 0.0000000715