PERMS AND COMBS difficult question (IMO) (1 Viewer)

pokemonlv10 · Sep 9, 2016

Hey guys, need help trying to solve this perms and combs question where:

A total of 5 marbles is selected at random from 4 differently coloured jars. Ear jar contains ten marbles, each being the same colour as the jar it is in and numbered from 1 to 10.

i) What is the probability that of the 5 marbles selected, three are numbered '4' and two are numbered '7'?

ii) Find the probability that the five selected marbles contain at least 4 marbles from the same jar.

Answer:1/27417
ii) 28/703

InteGrand · Sep 10, 2016

pokemonlv10 said:
Hey guys, need help trying to solve this perms and combs question where:

A total of 5 marbles is selected at random from 4 differently coloured jars. Ear jar contains ten marbles, each being the same colour as the jar it is in and numbered from 1 to 10.

i) What is the probability that of the 5 marbles selected, three are numbered '4' and two are numbered '7'?

ii) Find the probability that the five selected marbles contain at least 4 marbles from the same jar.

Answer:1/27417
ii) 28/703

$\noindent Note that the sample size is $\binom{40}{5}$.$

$\noindent i) We can choose the three $4$'s in $\binom{4}{3}$ ways. We can choose the two $7$'s in $\binom{4}{2}$ ways. So the number of ways to obtain three $4$'s and two $7$'s is $\binom{4}{3}\times \binom{4}{2}$. So the required probability is $\frac{\binom{4}{3}\times \binom{4}{2}}{\binom{40}{5}} = \frac{1}{27417}$.$

$\noindent ii) We can choose either exactly four marbles from one jar, or exactly five from one jar.$

$\noindent If we choose exactly four marbles from a jar, we can pick the jar to choose four from in $4$ ways. Once this jar is selected, we can pick the four marbles from it in $\binom{10}{4}$ ways, and the last marble in $30$ ways (one from the other jars). So $4\times \binom{10}{4} \times 30$ ways in this case.$

$\noindent If instead we choose five marbles from the same jar, we can choose the jar in $4$ ways and then choose the marbles in $\binom{10}{5}$ ways. So $4\times \binom{10}{5}$ ways in this case.$

$\noindent So the desired probability is $\frac{4\times \binom{10}{4} \times 30 + 4\times \binom{10}{5}}{\binom{40}{5}} = \frac{28}{703}$.$

pokemonlv10 · Sep 10, 2016

Thanks so much! i found that the sample size was 40C5 but after that I just couldnt work it out. Thank you!

PERMS AND COMBS difficult question (IMO) (1 Viewer)

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InteGrand

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