IMAT 2011 Q76 [Probability]

David has two boxes containing shapes.

In box A there are 4 stars and 2 hearts.

In box B there are 2 stars and 1 heart.

David takes, at random, a shape from box A and puts it into box B.

He then takes a shape from box B.

What is the probability that this shape is a star?

A. {\frac{1}{12}}

B. {\frac{4}{9}}

C. {\frac{2}{3}}

D. {\frac{3}{4}}

E. {\frac{4}{3}}

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What we can do is divide this question into two scenarios, keep in mind they need to end in David picking a star from Box B- so it narrows things down.

The probability David takes a heart out of Box A and inserts it into Box B is 1/3.

The probability he takes a star out of Box A and inserts it into Box B is 2/3.

We can now venture into two possibilities:

Scenario 1:
David takes a heart from Box A and inserts it into Box B.
David then picks a star from Box B.

We can now visualize that the probability he picks a inserts a heart is 1/3 and then picks a star would be 1/2.

We can permute this total scenario to be: \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}

Scenario 2:
David takes a star from Box A and inserts it into Box B.
David then picks a star from Box B.

We can now visualize that the probability he picks a inserts a star is 2/3 and then picks a star would be 3/4.

We can permute this total scenario to be: \frac{2}{3} \cdot \frac{3}{4} = \frac{1}{2}

Our final step is to add these probabilities to formulate the possibility of him picking a star in general, which means to add these two scenarios together:

\frac{1}{6} + \frac{1}{2} = \frac{2}{3}

Leaving our answer to be C, 2/3.