MATHEMATICS COLLEGE

Consider three boxes with numbered balls in them. Box A con- tains six balls numbered 1, . . . , 6. Box B contains twelve balls numbered 1, . . . , 12. Finally, box C contains four balls numbered 1, . . . , 4. One ball is selected from each urn uniformly at random. (a) What is the probability that the ball chosen from box A is labeled 1 if exactly two balls numbered 1 were selected
(b) What is the probability that the ball chosen from box B is 12 if the arithmetic mean of the three balls selected is exactly 7?

Answers

Answer 1

Answer:

Answer:

a) 0.73684

b) 2/3

Step-by-step explanation:

part a)

$P ( A is 1 / exactly two balls are 1) = (P ( A is 1 and that exactly two balls are 1))/(P (Exactly two balls are one))$

Using conditional probability as above:

(A,B,C)

Cases for numerator when:

P( A is 1 and exactly two balls are 1) = P( 1, not 1, 1) + P(1, 1, not 1)

= $((1)/(6)* (11)/(12)*(1)/(4)) + ((1)/(6)*(1)/(12)*(3)/(4)) = 0.048611111$

Cases for denominator when:

P( Exactly two balls are 1) = P( 1, not 1, 1) + P(1, 1, not 1) + P(not 1, 1 , 1)

$= ((1)/(6)* (11)/(12)*(1)/(4)) + ((1)/(6)*(1)/(12)*(3)/(4)) + ((5)/(6)*(1)/(12)*(1)/(4))= 0.0659722222$

Hence,

$P ( A is 1 / exactly two balls are 1) = (P ( A is 1 and that exactly two balls are 1))/(P (Exactly two balls are one)) = (0.048611111)/(0.06597222) \n\n= 0.73684$

Part b

$P ( B = 12 / A+B+C = 21) = (P ( B = 12 and A+B+C = 21))/(P (A+B+C = 21))$

Cases for denominator when:

P ( A + B + C = 21) = P(5,12,4) + P(6,11,4) + P(6,12,3)

$= 3*P(5,12,4 ) =3* (1)/(6)*(1)/(12)*(1)/(4)\n\n= (1)/(96)$

Cases for numerator when:

P (B = 12 & A + B + C = 21) = P(5,12,4) + P(6,12,3)

$= 2*P(5,12,4 ) =2* (1)/(6)*(1)/(12)*(1)/(4)\n\n= (1)/(144)$

Hence,

$P ( B = 12 / A+B+C = 21) = ((1)/(144) )/((1)/(96) )\n\n= (2)/(3)$

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