Question

8. There are three urns. Urnl contains 2 white and 3 black marbles, Urn2 contains 3 white and 2 black marbles, and Urn3 contains 1 white and 4 black marbles. First, the urn is selected equally likely, than a ball is drawn from the urn. a) We know that a white marble is chosen. What is the probability that the marble came from Urn2? b) We know that a black marble is chosen. What is the probability that the marble came from Urn2? 9. Assume that there is a football team which has 0.6 probability of win, 0.3 probability of lost, and 0.1 probability of drawn. This team plays three matches. a) What is the probability of the team wins at least two times and never lost? b) What is the probability of one time win, one time lost, and one time drawn? 10. Information bits {0,1} are sent over binary symmetric communication channel with conditional probabilities P(YX) as shown below. The priory probabilities of 0 and 1 are P(X=0)=0.3, P(X=1)=0.7. The error probability e=0.2. transmitter x 0 1 € receiver Y 0 E E 1-€ 1 a) If 1 is transmitted, what are the probabilities of receiving 0 and 1? P(Y=0X=1) and P(Y=1X=1) b) If 0 is received, what are the probabilities that 0 and 1 information bit is transmitted? P(X=0 Y=0) and P(X=1/Y=0)

Answer 1

8. a) Since all the urns are equally likely to be selected, so the probability of selecting an urn is = 1/3.

The probability of selecting a white marble from urn1 is = 1/3 * 2/5 = 2/15

The probability of selecting a white marble from urn2 is = 1/3 * 3/5 = 3/15

The probability of selecting a white marble from urn3 is = 1/3 * 1/5 = 1/15

The total probability of selecting a white marble is

= 2/15 + 3/15 + 1/15

= 6/15

= 2/5

The probability that the marble came from urn2 is, given that it is white is

$= \frac{3/15}{2/5} = 1/2$

b) The total probability of selecting a black marble is = 1 - 2/5 = 3/5

The probability of selecting a black marble from urn2 is = 1/3 * 2/5 = 2/15

The probability that the marble came from urn2 is, given that it is black is

$\bg_white = \frac{2/15}{3/5} = \frac{2}{9}$

9. Lets denote win, lost and draw by W, L and D.

a) The different cases in which the team wins at least two times and never loses is

i) WWD - P = 0.6 * 0.6 * 0.1 = 0.036

ii) WDW - P = 0.6 * 0.1 * 0.6 = 0.036

iii) DWW - P = 0.1 * 0.6 * 0.6 = 0.036

iv)WWW - P = 0.6 * 0.6 * 0.6 = 0.216

The total probability that the team wins at least two times and never loses is

= 0.036 + 0.036 + 0.036 + 0.216

= 0.384

b) The probability of one win, one loss and one draw is = 0.6 * 0.3 * 0.1 = 0.018

The number of ways in which the team can win once, lose and once and draw once out of the 3 games is

= 3! = 6

They are

WLD

WDL

LDW

LWD

DWL

DLW

The total probability that the team wins once, loses once and draws once is

= 0.018 * 6 = 0.108

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