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Mathematics 19 Online
OpenStudy (anonymous):

the probability that it rains on any day in a particular month is p. The probability that it rains on only one day on two consecutive days in the month is 0.32 Find the possible values of p

OpenStudy (anonymous):

ok that was wrong probably \(p+p-2p^2=.32\) since only rains one day

OpenStudy (anonymous):

still a very bad assumption, it is unlikely that the events are independent

OpenStudy (anonymous):

The answers are : 0.4 , 0.8 i have no idea how to start this question..

OpenStudy (anonymous):

@satellite73 p + p - 2p^2 = .32 can you explain me how you got this info ?

OpenStudy (anonymous):

\[p(1-p)+p(1-p)=0.32\] if the events are independent, and it rains only one day, the probability of this happening is- either of these, 1) rain(p) and no rain (1-p) 2) no-rain(1-p) and rain(p)

OpenStudy (anonymous):

but htis gives 0.2 and 0.8

OpenStudy (anonymous):

thank you i try to work it out for 30 mins and obtain the answer...

OpenStudy (anonymous):

you say your answers are 0.4 and 0.8? \[2p(1-p)=0.32\implies p-p^2=0.16\\p^2-p+0.16=0\\ p={1\pm\sqrt{1-4(0.16)}\over2}\]

OpenStudy (anonymous):

\[p=0.5\pm0.3\]

OpenStudy (anonymous):

p^2 - p + 0.16 = 0 \[(p -1/2)^2 = 0.09 \] Complete square P - 0.5 = 0.3 or P - 0.5 = -.03 P = 0.8 , 0.2

OpenStudy (anonymous):

you used the formula i made it by completing square both is great :) we have 2 methods now to do that cool :D !

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