Question

Box A contains 3 red and 2 blue marbles while Box B contains 2 red and 8 blue marbles. A fair coin is tossed. If the coin turns up heads, a marble is chosen from Box A; if it turns up tails, a marble is chosen from Box B.
A. Find the probability that a red marble is chosen.
B. Referring to above information, suppose that the person who tosses the coin does not reveal whether it has turned up heads or tails (so that the box from which a marble was chosen is not revealed) but does reveal that a red marble was chosen. What is the probability that Box A was chosen (i.e., the coin turned up heads)?

Answer 1

Answer:

A. $\frac{4}{10}$

B. $\frac{3}{4}$

Step-by-step explanation:

Given - Box A contains 3 red and 2 blue marbles while Box B contains 2

            red and 8 blue marbles. A fair coin is tossed. If the coin turns up

            heads, a marble is chosen from Box A; if it turns up tails, a marble

            is chosen from Box B.

To find -  A. Find the probability that a red marble is chosen.

              B. Referring to above information, suppose that the person who

                  tosses the coin does not reveal whether it has turned up

                  heads or tails (so that the box from which a marble was

                  chosen is not revealed) but does reveal that a red marble was

                  chosen. What is the probability that Box A was chosen (i.e.,

                  the coin turned up heads)?

Proof -

As given ,

Box A : 3 red marbles and 2 blue marbles.

Box B : 2 red marbles and 8 blue marbles.

a.)

As we have to find the probability that a red marble is chosen.

A red marble is chosen in two different ways - either it can be from Box A or Box be .

If Box A is chosen , then Probability of red ball is $\frac{3}{5}$

If Box B is chosen , Then Probability of Red ball is $\frac{2}{10} = \frac{1}{5}$

Now,

Probability of Box A is Chosen = $\frac{1}{2}$

Probability of Box B is chosen = $\frac{1}{2}$

∴ we get

Probability that a Red marble is chosen =  $\frac{1}{2}$×$\frac{3}{5}$ + $\frac{1}{2}$×$\frac{1}{5}$ = $\frac{3}{10} + \frac{1}{10} = \frac{4}{10}$ = 0.4

b.)

Now,

Probability that Box A is chosen given that Red marble is chosen is

P(Box A | Red marble ) = $\frac{(\frac{3}{5})(\frac{1}{2}) }{(\frac{3}{5})(\frac{1}{2}) + (\frac{1}{5})(\frac{1}{2})} = \frac{\frac{3}{10} }{\frac{3}{10} + \frac{1}{10} } = \frac{\frac{3}{10} }{\frac{4}{10}}$ = $\frac{3}{4}$