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प्रश्न
Suppose we have four boxes A, B, C, D containing coloured marbles as given below:
Figure
One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A? box B? box C?
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उत्तर
Let R be the event of drawing the red marble.
Let EA, EB and EC denote the events of selecting box A, box B and box C, respectively.
Total number of marbles = 40
Number of red marbles = 15
\[\therefore P\left( R \right) = \frac{15}{40} = \frac{3}{8}\]
Probability of drawing a red marble from box A is given by P(EA/R).
Probability of drawing a red marble from box B is given by P(EB/R).
\[\therefore P\left( E_B /R \right) = \frac{P\left( E_B \cap R \right)}{P\left( R \right)} = \frac{\frac{6}{40}}{\frac{3}{8}} = \frac{2}{5}\]
Probability of drawing a red marble from box C is given by P(EC/R)
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Solution: Let A, C and T be the events that Mr. X goes to office by Auto, Car and Train respectively. Let L be event that he is late.
Given that P(A) = `square`, P(C) = `square`
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P(L) = P(A ∩ L) + P(C ∩ L) + P(T ∩ L)
`="P"("A")*"P"("L"//"A") + "P"("C")*"P"("L"//"C") + "P"("T")*"P"("L"//"T")`
`= square * square + square * square + square * square`
`= square + square + square`
`= square`
`"P"("C"//"L") = ("P"("L" ∩ "C"))/("P"("L"))`
= `("P"("C") * "P"("L"//"C"))/("P"("L"))`
`= (square * square)/square`
`= square`
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