Menu Close

Question-180576




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
✓
$$\checkmark \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *