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Question Number 180576 by Acem last updated on 14/Nov/22

Commented by Acem last updated on 14/Nov/22

a• What′s the probability that Rosy would stand on the far left and Nathalie on the far right? b• Same question but and you stand in the middle in case you were 11 person? c• Same question ′a′ but or you stand on the 2nd place on the right in case you were 11 too? d• What′s the probability that they stand look like the picture above?

$${a}\bullet\:{What}'{s}\:{the}\:{probability}\:{that}\:{Rosy}\:{would}\:{stand} \\ $$$$\:{on}\:{the}\:{far}\:{left}\:{and}\:{Nathalie}\:{on}\:{the}\:{far}\:{right}? \\ $$$${b}\bullet\:{Same}\:{question}\:{but}\:{and}\:{you}\:{stand}\:{in}\:{the}\:{middle} \\ $$$$\:{in}\:{case}\:{you}\:{were}\:\mathrm{11}\:{person}? \\ $$$${c}\bullet\:{Same}\:{question}\:'{a}'\:{but}\:{or}\:{you}\:{stand}\:{on}\:{the}\:\mathrm{2}{nd} \\ $$$$\:{place}\:{on}\:{the}\:{right}\:{in}\:{case}\:{you}\:{were}\:\mathrm{11}\:{too}? \\ $$$${d}\bullet\:{What}'{s}\:{the}\:{probability}\:{that}\:{they}\:{stand}\:{look}\:{like} \\ $$$$\:{the}\:{picture}\:{above}? \\ $$$$ \\ $$

Answered by mr W last updated on 14/Nov/22

a) p=((5!)/(7!))=(1/(42))

$$\left.{a}\right) \\ $$$${p}=\frac{\mathrm{5}!}{\mathrm{7}!}=\frac{\mathrm{1}}{\mathrm{42}} \\ $$

Commented by Acem last updated on 14/Nov/22

✓

$$\checkmark \\ $$

Answered by mr W last updated on 14/Nov/22

b) p=((10!)/(11!))=(1/(11))

$$\left.{b}\right) \\ $$$${p}=\frac{\mathrm{10}!}{\mathrm{11}!}=\frac{\mathrm{1}}{\mathrm{11}} \\ $$

Commented by Acem last updated on 14/Nov/22

✓

$$\checkmark \\ $$

Commented by Acem last updated on 14/Nov/22

Sir, same question ′a′ , Rosy, Nathalie and you

$${Sir},\:{same}\:{question}\:'{a}'\:,\:{Rosy},\:{Nathalie}\:{and}\:{you} \\ $$

Commented by mr W last updated on 14/Nov/22

then p=((8!)/(11!))=(1/(990))

$${then} \\ $$$${p}=\frac{\mathrm{8}!}{\mathrm{11}!}=\frac{\mathrm{1}}{\mathrm{990}} \\ $$

Answered by mr W last updated on 14/Nov/22

c) p=((8!)/(11!))=(1/(990))

$$\left.{c}\right) \\ $$$${p}=\frac{\mathrm{8}!}{\mathrm{11}!}=\frac{\mathrm{1}}{\mathrm{990}} \\ $$

Commented by Acem last updated on 14/Nov/22

Same note above but Or you

$${Same}\:{note}\:{above}\:{but}\:{Or}\:{you} \\ $$

Commented by mr W last updated on 14/Nov/22

then please make your question clear enough. “but and you”, “but or you”, what do you mean exactly?

$${then}\:{please}\:{make}\:{your}\:{question}\:{clear}\: \\ $$$${enough}.\:``{but}\:{and}\:{you}'',\:``{but}\:{or}\:{you}'', \\ $$$${what}\:{do}\:{you}\:{mean}\:{exactly}? \\ $$

Commented by Acem last updated on 14/Nov/22

Hello friend, b• Same ′a′ , in case you were 11: I meant that Rosy would stand on the far left and Nathalie on the far right and you stand in the middle in case you were 11 person? c• Rosy would stand on the far left and Nathalie on the far right or you stand in the 2nd place on the right in case you were 11 person?

$${Hello}\:{friend}, \\ $$$${b}\bullet\:{Same}\:'{a}'\:,\:{in}\:{case}\:{you}\:{were}\:\mathrm{11}: \\ $$$$\:{I}\:{meant}\:{that}\:{Rosy}\:{would}\:{stand}\:{on}\:{the}\:{far}\:{left} \\ $$$$\:{and}\:{Nathalie}\:{on}\:{the}\:{far}\:{right}\:{and}\:{you}\:{stand}\:{in} \\ $$$$\:{the}\:{middle}\:{in}\:{case}\:{you}\:{were}\:\mathrm{11}\:{person}? \\ $$$$ \\ $$$${c}\bullet\:{Rosy}\:{would}\:{stand}\:{on}\:{the}\:{far}\:{left}\:{and}\:{Nathalie} \\ $$$$\:{on}\:{the}\:{far}\:{right}\:{or}\:{you}\:{stand}\:{in}\:{the}\:\mathrm{2}{nd}\:{place} \\ $$$$\:{on}\:{the}\:{right}\:{in}\:{case}\:{you}\:{were}\:\mathrm{11}\:{person}? \\ $$$$ \\ $$

Answered by mr W last updated on 14/Nov/22

d) p=(1/(7!))=(1/(5040))

$$\left.{d}\right) \\ $$$${p}=\frac{\mathrm{1}}{\mathrm{7}!}=\frac{\mathrm{1}}{\mathrm{5040}} \\ $$

Commented by Acem last updated on 14/Nov/22

✓

$$\checkmark \\ $$

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