# Out of three bags that each contains 10 marbles : Bag 1 has 6 red and 4 blue marbles Bag 2 has 5 red and 5 blue marbles. Bag 3 has 2 red and 8 blue marbles. What is the Probability of choosing a red marble? Option 1) $\frac{11}{300}$ Option 2) $\frac{13}{300}$ Option 3) $\frac{17}{300}$ Option 4) $\frac{7}{300}$

As we learned

The law of Total Probability -

Let S be the sample space and E1, E2, ......En be n mutually exclusive and exhaustive events associated with a random experiment.

- wherein

$\dpi{100} \Rightarrow P\left ( A \right )= P\left ( A\cap E_{1} \right )+P\left ( A\cap E_{2} \right )+\cdots +P\left ( A\cap E_{n} \right )$

$\dpi{100} \Rightarrow P\left ( A \right )= P\left ( E_{1} \right )\cdot P\left ( \frac{A}{E_{1}} \right )+P\left ( E_{2} \right )\cdot P\left ( \frac{A}{E_{2}} \right )+\cdots P\left ( E_{n} \right )\cdot P\left ( \frac{A}{E_{n}} \right )$

where A is any event which occurs with E1, E2, E3......En.

$P\left ( \frac{R}{B_{1}} \right )=0.6\: ;\: P\left ( \frac{R}{B_{2}} \right )=0.2$

Thus $P\left ( R \right )=\frac{1}{3}\left ( 0.6 \right )+\frac{1}{3}\left ( 0.5 \right )+\frac{1}{3}\left ( 0.2 \right )$

$=\frac{0.11}{3}$

$=\frac{11}{300}$

Option 1)

$\frac{11}{300}$

Option 2)

$\frac{13}{300}$

Option 3)

$\frac{17}{300}$

Option 4)

$\frac{7}{300}$

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