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Question Number 4753 by FilupSmith last updated on 05/Mar/16

∗  ∗  ∗  ∗  ∗  ∗  ∗  ∗  ∗  ∗  ∗  ∗  ∗  ∗  ∗    Selecting a random star is a  (1/(15)) chance  at random. Lets say you have to pick  a second random star that is next to it.  Either above, below, or to the side.  What this the probability of selecting  the two stars correctly?    −−−−−−    As far as i have worked out, if the first  star is not an edge or corner, the second  has a (1/4) chance    if the first is an edge, the second had a  (1/3) chance    if the first is a corner, the second is a (1/2)  chance.    what is the overall probability???

$$\ast\:\:\ast\:\:\ast\:\:\ast\:\:\ast \\ $$$$\ast\:\:\ast\:\:\ast\:\:\ast\:\:\ast \\ $$$$\ast\:\:\ast\:\:\ast\:\:\ast\:\:\ast \\ $$$$ \\ $$$$\mathrm{Selecting}\:\mathrm{a}\:\mathrm{random}\:\mathrm{star}\:\mathrm{is}\:\mathrm{a}\:\:\frac{\mathrm{1}}{\mathrm{15}}\:\mathrm{chance} \\ $$$$\mathrm{at}\:\mathrm{random}.\:\mathrm{Lets}\:\mathrm{say}\:\mathrm{you}\:\mathrm{have}\:\mathrm{to}\:\mathrm{pick} \\ $$$$\mathrm{a}\:\mathrm{second}\:\mathrm{random}\:\mathrm{star}\:\mathrm{that}\:\mathrm{is}\:\mathrm{next}\:\mathrm{to}\:\mathrm{it}. \\ $$$${Either}\:{above},\:{below},\:{or}\:{to}\:{the}\:{side}. \\ $$$$\mathrm{What}\:\mathrm{this}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{of}\:\mathrm{selecting} \\ $$$$\mathrm{the}\:\mathrm{two}\:\mathrm{stars}\:\mathrm{correctly}? \\ $$$$ \\ $$$$−−−−−− \\ $$$$ \\ $$$$\mathrm{As}\:\mathrm{far}\:\mathrm{as}\:\mathrm{i}\:\mathrm{have}\:\mathrm{worked}\:\mathrm{out},\:\mathrm{if}\:\mathrm{the}\:\mathrm{first} \\ $$$$\mathrm{star}\:\mathrm{is}\:\mathrm{not}\:\mathrm{an}\:\mathrm{edge}\:\mathrm{or}\:\mathrm{corner},\:\mathrm{the}\:\mathrm{second} \\ $$$$\mathrm{has}\:\mathrm{a}\:\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{chance} \\ $$$$ \\ $$$$\mathrm{if}\:\mathrm{the}\:\mathrm{first}\:\mathrm{is}\:\mathrm{an}\:\mathrm{edge},\:\mathrm{the}\:\mathrm{second}\:\mathrm{had}\:\mathrm{a} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{chance} \\ $$$$ \\ $$$$\mathrm{if}\:\mathrm{the}\:\mathrm{first}\:\mathrm{is}\:\mathrm{a}\:\mathrm{corner},\:\mathrm{the}\:\mathrm{second}\:\mathrm{is}\:\mathrm{a}\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{chance}. \\ $$$$ \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{overall}\:\mathrm{probability}??? \\ $$$$ \\ $$

Commented by prakash jain last updated on 05/Mar/16

If the first star is not an edge or corner then  you have 4 possibilities out of 14 remaining  how did you come with (1/4)?

$$\mathrm{If}\:\mathrm{the}\:\mathrm{first}\:\mathrm{star}\:\mathrm{is}\:\mathrm{not}\:\mathrm{an}\:\mathrm{edge}\:\mathrm{or}\:\mathrm{corner}\:\mathrm{then} \\ $$$$\mathrm{you}\:\mathrm{have}\:\mathrm{4}\:\mathrm{possibilities}\:\mathrm{out}\:\mathrm{of}\:\mathrm{14}\:\mathrm{remaining} \\ $$$$\mathrm{how}\:\mathrm{did}\:\mathrm{you}\:\mathrm{come}\:\mathrm{with}\:\frac{\mathrm{1}}{\mathrm{4}}? \\ $$

Commented by FilupSmith last updated on 05/Mar/16

Oh, sorry i meant there is a (1/4) chance ti  get the second in relation to the first

$$\mathrm{Oh},\:{sorry}\:\mathrm{i}\:\mathrm{meant}\:\mathrm{there}\:\mathrm{is}\:\mathrm{a}\:\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{chance}\:\mathrm{ti} \\ $$$$\mathrm{get}\:\mathrm{the}\:\mathrm{second}\:\mathrm{in}\:\mathrm{relation}\:\mathrm{to}\:\mathrm{the}\:\mathrm{first} \\ $$

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