Parcels from sender S to receiver R pass sequentially through two post-offices. Each post-office has a probability \(\frac{1}{5}\) of losing an incoming parcel, independently of all other parcels. Given that a parcel is lost, the probability that it was lost by the second post-office is_________.
Concept:
\(p\left( {\bar A} \right) = 1 - p\left( A \right)\)
Conditional Probability: The probability of an event A given that event B has occurred is defined as:
\(p\left( {\frac{A}{B}} \right) = \frac{{p\left( {A\; \cap B} \right)}}{{p\left( B \right)}}\)
Calculation:
Probability of losing a parcel = 1/5
Probability of successfully passing a parcel = 1 - Probability of losing a parcel
= 1-1/5 = 4/5
There are two cases for losing the parcel from sender S to receiver R.
Lost at first post-office:
Probability of losing the parcel at first post-office \( = \frac{1}{5}\)
Successfully passed at first post-office but lost at 2nd post-office:
Probability of losing the parcel corresponding to this case will be:
\( = \frac{4}{5} \times \frac{1}{5}\)
Net probability of losing the parcel will be:
\( = \frac{1}{5} + \frac{4}{5} \times \frac{1}{5} = \frac{9}{{25}}\)
\(p\left( {\frac{{Parcel\;lost\;at\;2nd\;post\;office}}{{parcel\;lost}}} \right)\)
\( = \frac{{p\left( {Parcel\;lost\;at\;2nd\;post\;office\; \cap \;parcel\;lost} \right)}}{{p\left( {parcel\;lost} \right)}}\)
\( = \;\frac{{p\left( {parcel\;lost\;at\;2nd\;post - office} \right)}}{{p\left( {parcel\;lost} \right)}}\)
\( = \frac{{4/25}}{{9/25}} = \frac{4}{9} \approx 0.44\)
Let A and B be two events such that \(P(\overline{A \cup B}) = \dfrac{1}{6}\) , P(A ∩ B) = \(\dfrac{1}{4}\) and P(A̅) = \(\dfrac{1}{4}\) , where A̅ stands for complement of event A. Then, events A and B are:
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In a bulb factory, machines P, Q and R manufacture respectively 25%, 35% and 40% of the total. Of their output 5, 4 and 2 percent respectively are defective bulbs. A bulb is drawn at random and it is found to be defective. What is the probability that it was manufactured by machine Q?