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3 years ago

Plz help!!!!

1 answer:

3 0

Time difference between 10:53AM and 10:53PM same day = 12 hours.
Time difference between 10:53PM and 12 midnight (next day) = 1 hour 7 minutes. 
Time difference between 12 midnight and 1:05 AM (next day) = 1 hour 5 minutes. 
Time difference between 1:05 AM (next day) and 1:05 PM (next day) = 12 hours. 

Time difference between 10:53 AM one day and 1:05 PM the next day = 26 hours 12 minutes. Half of that is 13 hours 6 minutes. 

10:53 AM + 12 hours = 10:53 PM. 
10:53PM + 1 hour 6 mins = 11:59 PM

My phone rang midway between 10:53 AM one day and 1:05 PM the next day. It rang at 11:59 PM.

i hope this helps! :)

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We're told that

$P(A\cap B)=0.15$

$P(A\cup B)^C=0.06\implies P(A\cup B)=0.94$

$P(B\mid A)=P(B^C\mid A)=0.5$

where the last fact is due to the law of total probability:

$P(A)=P(A\cap B)+P(A\cap B^C)$

$\implies P(A)=P(B\mid A)P(A)+P(B^C\mid A)P(A)$

$\implies 1=P(B\mid A)+P(B^C\mid A)$

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By definition of conditional probability, we have

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$\implies\dfrac{P(A\cap B)}{P(A)}=\dfrac{P(A\cap B^C)}{P(A)}$

$\implies P(A\cap B)=P(A\cap B^C)$

We make use of the addition rule and complementary probabilities to rewrite this as

$P(A\cap B)=P(A\cap B^C)$

$\implies P(A)+P(B)-P(A\cup B)=P(A)+P(B^C)-P(A\cup B^C)$

$\implies P(B)-[1-P(A\cup B)^C]=[1-P(B)]-P(A\cup B^C)$

$\implies2P(B)=2-[P(A\cup B)^C+P(A\cup B^C)]$

$\implies2P(B)=[1-P(A\cup B)^C]+[1-P(A\cup B^C)]$

$\implies2P(B)=P(A\cup B)+P(A\cup B^C)^C$

$\implies2P(B)=P(A\cup B)+P(A^C\cap B)\quad(*)$

By the law of total probability,

$P(B)=P(A\cap B)+P(A^C\cap B)$

$\implies P(A^C\cap B)=P(B)-P(A\cap B)$

and substituting this into $(*)$ gives

$2P(B)=P(A\cup B)+[P(B)-P(A\cap B)]$

$\implies P(B)=P(A\cup B)-P(A\cap B)$

$\implies P(B)=0.94-0.15=\boxed{0.79}$

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