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Question Number 131062 by benjo_mathlover last updated on 01/Feb/21

You flip two fair six−sidded dice and add  the number of spots. what probability of getting  a number divisible by 2 but not by 5 ?

$$\mathrm{You}\:\mathrm{flip}\:\mathrm{two}\:\mathrm{fair}\:\mathrm{six}−\mathrm{sidded}\:\mathrm{dice}\:\mathrm{and}\:\mathrm{add} \\ $$$$\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{spots}.\:\mathrm{what}\:\mathrm{probability}\:\mathrm{of}\:\mathrm{getting} \\ $$$$\mathrm{a}\:\mathrm{number}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{2}\:\mathrm{but}\:\mathrm{not}\:\mathrm{by}\:\mathrm{5}\:? \\ $$

Answered by EDWIN88 last updated on 01/Feb/21

Write L_n  for even the number is  divisible by n. Now P(L_2 )=1/2.   (count the case; or more elegantly notice  that each die has the same number of  odd and even faces and work from there)  Now P(L_2 −L_5 )= P(L_2 )−P(L_2 ∩L_5 )  but L_2 ∩L_5  contains only three  outcomes (6,4),(5,5) and (4,6) so   P(L_2 −L_5 )= (1/2)−(3/(36))=(5/(12))

$${Write}\:{L}_{{n}} \:{for}\:{even}\:{the}\:{number}\:{is} \\ $$$${divisible}\:{by}\:{n}.\:{Now}\:{P}\left({L}_{\mathrm{2}} \right)=\mathrm{1}/\mathrm{2}.\: \\ $$$$\left({count}\:{the}\:{case};\:{or}\:{more}\:{elegantly}\:{notice}\right. \\ $$$${that}\:{each}\:{die}\:{has}\:{the}\:{same}\:{number}\:{of} \\ $$$$\left.{odd}\:{and}\:{even}\:{faces}\:{and}\:{work}\:{from}\:{there}\right) \\ $$$${Now}\:{P}\left({L}_{\mathrm{2}} −{L}_{\mathrm{5}} \right)=\:{P}\left({L}_{\mathrm{2}} \right)−{P}\left({L}_{\mathrm{2}} \cap{L}_{\mathrm{5}} \right) \\ $$$${but}\:{L}_{\mathrm{2}} \cap{L}_{\mathrm{5}} \:{contains}\:{only}\:{three} \\ $$$${outcomes}\:\left(\mathrm{6},\mathrm{4}\right),\left(\mathrm{5},\mathrm{5}\right)\:{and}\:\left(\mathrm{4},\mathrm{6}\right)\:{so}\: \\ $$$${P}\left({L}_{\mathrm{2}} −{L}_{\mathrm{5}} \right)=\:\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{36}}=\frac{\mathrm{5}}{\mathrm{12}} \\ $$

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