Question

Each of the 2 balls is painted either black or gold and then placed in an urn. Suppose that each ball is colored black with probability$\frac{1}{2}$and that these events are independent . (a) Suppose that you obtain information that the gold paint has been used (and thus at least one of the balls is painted gold). Compute the conditional probability that both balls are painted gold. (b) Suppose now that the urn tips over and 1 ball falls out. It is painted gold. What is the probability that both balls are gold in this case? Explain

Short answer

The probability that both balls are painted gold provided at least one of the balls is painted gold is$\frac{1}{3}$

The probability of the second gold provided that the first ball ( fall out ball from the urn) is gold is$\frac{1}{2}$

Step-by-step solution

Given Information of (a)

Each of the 2 balls is painted either black or gold and then placed in an urn.

Each ball is colored black with probability $\frac{1}{2}$ and these events are independent

We have to compute the conditional probability that both balls are painted gold.

Explanation of (a)

From the given data, the probability of painting the ball with black or gold is $\frac{1}{2}$and the events are independent.

Assume$E_1$is the event that both balls painted with black, $E_2$is the event that both ball painted with Gold and $E_3$is the event that one ball painted with black and another one is painted with gold doesn't matter with the order.

Therefore,

$P\left(E_1\right)=\frac{1}{2}\times \frac{1}{2}$

$=\frac{1}{4}$

$P\left(E_2\right)=\frac{1}{2}\times \frac{1}{2}$

$=\frac{1}{4}$

$P\left(E_3\right)=2\times \frac{1}{2}\times \frac{1}{2}\left(\begin{array}{l}\because \text{order doesn't matter}\\ P\left(BG\right)+P\left(GB\right)\end{array}\right)$

$=\frac{1}{2}$

Probability of at least one of the balls painted gold  

Suppose the gold paint has been used ( at least one of the balls is painted). Then need to find the conditional probability that both balls are painted gold.

Probability of at least one of the balls painted gold is

$P\left(\text{at least one gold}\right)=P\left(E_2\right)+P\left(E_3\right)$

$=\frac{1}{4}+\frac{1}{2}$

$=\frac{3}{4}$

Probability of at least one of the balls painted gold

$\frac{3}{4}$

Final Answer (a)

The probability that both balls are painted gold provided at least one of the balls is painted gold is

$P\left(E_2\mid \text{at least one gold}\right)=\frac{P\left(E_2\cap \text{at least one gold}\right)}{P\left(\text{at least one gold}\right)}$

$=\frac{P\left(E_2\right)}{\frac{3}{4}}$

$=\frac{1}{4}\times \frac{4}{3}$

$=\frac{1}{3}$

Step % Given Information (b)

Each of the 2 balls is painted either black or gold and then placed in an urn.

Each ball is colored black with probability and these events are independent

We have to compute the probability that both balls are gold in this case if the urn tips over and 1 ball falls out. It is painted gold

Explanation of (b)

Suppose 1 ball falls out from the urn and know it was painted with gold. Then need to find the probability that both balls are gold. The probability of the second gold provided that the first ball ( fall out ball from the urn) is gold is

$P(\text{Second}\mathrm{G}\mid \text{First}\mathrm{G})=\frac{P(\text{Second}\mathrm{G}\cap \text{First}\mathrm{G})}{P\left(\text{First}\mathrm{G}\right)}$

$=\frac{P\left(\text{Second}G\right)P\left(\text{First}G\right)}{P\left(\text{First}G\right)}\left(\begin{array}{l}\text{" events are}\\ \text{independent}\end{array}\right)$

$=\frac{\frac{1}{2}\times \frac{1}{2}}{\frac{1}{2}}$

$=\frac{1}{2}$

Final Answer of (b)

The probability of the second gold provided that the first ball ( fall out ball from the urn) is gold $=\frac{1}{2}$