Q.3.24 - A First Course in Probability | StudyQuestionHub

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Q.3.24

Question

Each of $2$ 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.

Step-by-Step Solution

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Answer

a)The conditional probability that both balls are painted gold $\frac{1}{3} [2 e x]$

b) The probability that both balls are gold is $\frac{1}{2}$

1Step1:Introduction

Considered events:

$G_{1}$ -The first of the two balls is gold in color.

$G_{2}$ - The second of the two balls is gold in color.

Probabilities

$P (G_{1}) = \frac{1}{2} \Rightarrow P (G_{1}^{c}) = \frac{1}{2}$

$P (G_{2}) = \frac{1}{2} \Rightarrow P (G_{2}^{c}) = \frac{1}{2}$

The events $G_{1}$ and $G_{2}$ are independent

2Step2: Find P G 1 G 2 ∣ G 1 ∪ G 2

Begin by defining conditional probability.

$P (G_{1} G_{2} ∣ (G_{1} \cup G_{2})) = \frac{P [G_{1} G_{2} (G_{1} \cup G_{2})]}{P (G_{1} \cup G_{2})}$

We use this to change the definition (1),

$P [G_{1} G_{2} ∣ (G_{1} \cup G_{2})] = \frac{P (G_{1} G_{2})}{P [{(G_{1}^{c} G_{2}^{c})}^{c}]}$

$= \frac{P (G_{1}) P (G_{2})}{1 - P [(G_{1}^{c} G_{2}^{c})]}$

$= \frac{P (G_{1}) P (G_{2})}{1 - P (G_{1}^{c}) P (G_{2}^{c})}$

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

$= \frac{1}{3}$

3Step2:The probability that the other one is golden(part b)

What are the chances that the other ball will be golden as well?

Either $G_{1}$ happens, and the desired probability is $P (G_{2} ∣ G_{1})$

or $G_{2}$ happens, and the desired probability is $P (G_{1} ∣ G_{2})$ .

According to the definition of independence,

$\begin{aligned} P (G_{1} ∣ G_{2}) = P (G_{1}) = \frac{1}{2} \\ P (G_{2} ∣ G_{1}) = P (G_{2}) = \frac{1}{2} \end{aligned}$

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