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  • There are 4 white and 3 black balls in a box. In another box there are 3 white and 4 black balls. An unbiased dice is rolled. If it shows a number less than or equal to 3 then a ball is drawn from

There are 4 white and 3 black balls in a box. In another box there are 3 white and 4 black balls. An unbiased dice is rolled. If it shows a number less than or equal to 3 then a ball is drawn from the first box but if it shows a number more than 3 then a ball is drawn from the second box. If the ball drawn is black then the probability that the ball was drawn from the first box is

Option: 1

\frac{1}{2}


Option: 2

\frac{6}{7}


Option: 3

\frac{4}{7}


Option: 4

\frac{3}{7}


Answers (1)

best_answer

\mathrm{E_1=} the event of drawing from the first box.

\mathrm{E_2=} the event of drawing from the second box.

\mathrm{P\left(E_1\right)=\frac{3}{6}=\frac{1}{2} \quad \text { and } \quad P\left(E_2\right)=\frac{3}{6}=\frac{1}{2} .}

Clearly, \mathrm{E_1 \text { and } E_2 } are mutually exclusive and exhaustive.

Now, \mathrm{P\left(\frac{B}{E_1}\right)=\frac{3}{7} \text { and } P\left(\frac{B}{E_2}\right)=\frac{4}{7} }

We have to find \mathrm{P\left(\frac{E_1}{B}\right)}

By Bayes' theorem,

\mathrm{P\left(\frac{E_1}{B}\right)=\frac{P\left(E_1\right) \cdot P\left(\frac{B}{E_1}\right)}{P\left(E_1\right) \cdot P\left(\frac{B}{E_1}\right)+P\left(E_2\right) \cdot P\left(\frac{B}{E_2}\right)}=\frac{\frac{1}{2} \cdot \frac{3}{7}}{\frac{1}{2} \cdot \frac{3}{7}+\frac{1}{2} \cdot \frac{4}{7}}=\frac{3}{7}}

 

Posted by

Gaurav

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