Question

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?

Answer 1

Answer:

a) 0.73684

b) 2/3

Step-by-step explanation:

part a)

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

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)

= $(\frac{1}{6}* \frac{11}{12}*\frac{1}{4}) + (\frac{1}{6}*\frac{1}{12}*\frac{3}{4}) = 0.048611111$

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)

$= (\frac{1}{6}* \frac{11}{12}*\frac{1}{4}) + (\frac{1}{6}*\frac{1}{12}*\frac{3}{4}) + (\frac{5}{6}*\frac{1}{12}*\frac{1}{4})= 0.0659722222$

Hence,

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

Part b

$P ( B = 12 / A+B+C = 21) = \frac{P ( B = 12 and A+B+C = 21)}{P (A+B+C = 21)}$

Cases for denominator when:

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

$= 3*P(5,12,4 ) =3* \frac{1}{6}*\frac{1}{12}*\frac{1}{4}\\\\= \frac{1}{96}$

Cases for numerator when:

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

$= 2*P(5,12,4 ) =2* \frac{1}{6}*\frac{1}{12}*\frac{1}{4}\\\\= \frac{1}{144}$

Hence,

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