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Each of 2balls is painted either black or gold and then placed in an urn. Suppose that each ball is colored black with probability 12and 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

Expert verified

a)The conditional probability that both balls are painted gold 13[2ex]

b) The probability that both balls are gold is12

Step by step solution

01

Step1:Introduction

Considered events:

G1-The first of the two balls is gold in color.

G2- The second of the two balls is gold in color.

Probabilities

PG1=12PG1c=12

PG2=12PG2c=12

The events G1and G2are independent

02

FindPG1G2∣G1∪G2

Begin by defining conditional probability.

PG1G2G1G2=PG1G2G1G2PG1G2

We use this to change the definition (1),

localid="1649413957621" PG1G2G1G2=PG1G2PG1cG2cc

=PG1PG21PG1cG2c

=PG1PG21PG1cPG2c

=121211212

=13

03

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

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

Either G1 happens, and the desired probability isPG2G1

or G2happens, and the desired probability isPG1G2.

According to the definition of independence,

PG1G2=PG1=12PG2G1=PG2=12

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Most popular questions from this chapter

Consider an urn containing 12balls, which 8are white. A sample of size4is to be drawn with replacement (without replacement). What is the conditional probability (in each case) that the first and third balls drawn will be white given that the sample drawn contains exactly3 white balls?

(a) Prove that if Eand Fare mutually exclusive, then

localid="1647926638131" P(EEF)=P(E)P(E)+P(F)

(b) Prove that if localid="1647926673038" Ei,i1are mutually exclusive, then

localid="1648539605315" PEji=1Ei=PEji=1PEi

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