hw2sol

# hw2sol - UCSD ECE 153 Handout#11 Prof Young-Han Kim...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: UCSD ECE 153 Handout #11 Prof. Young-Han Kim Thursday, April 14, 2011 Solutions to Homework Set #2 (Prepared by TA Lele Wang) 1. Polya’s urn. Suppose we have an urn containing one red ball and one blue ball. We draw a ball at random from the urn. If it is red, we put the drawn ball plus another red ball into the urn. If it is blue, we put the drawn ball plus another blue ball into the urn. We then repeat this process. At the n-th stage, we draw a ball at random from the urn with n +1 balls, note its color, and put the drawn ball plus another ball of the same color into the urn. (a) Find the probability that the first ball is red. (b) Find the probability that the second ball is red. (c) Find the probability that the first three balls are all red. (d) Find the probability that two of the first three balls are red. Solution: Let X i denote the color of the i-th ball. (a) By symmetry, P { X 1 = R } = 1 / 2. (b) Again by symmetry, P { X i = R } = 1 / 2 for all i . Alternatively, by the law of total probability, we have P { X 2 = R } = P { X 1 = R } P { X 2 = R | X 1 = R } + P { X 1 = B } P { X 2 = R | X 1 = B } = 1 2 × 2 3 + 1 2 × 1 3 = 1 2 . (c) By the chain rule, we have P { X 1 = R,X 2 = R,X 3 = R } = P { X 1 = R } P { X 2 = R | X 1 = R } P { X 3 = R | X 2 = R,X 1 = R } = 1 2 × 2 3 × 3 4 = 1 4 . (d) Let N denote the number of red balls in the first three draws. From part (c), we know that P { N = 3 } = 1 / 4 = P { N = 0 } , where the latter identity follows by symmetry. Also we have P { N = 2 } = P { N = 1 } . Thus, P { N = 2 } must be 1 / 4. 1 Alternatively, we have P { N = 2 } = P { X 1 = B,X 2 = R,X 3 = R } + P { X 1 = R,X 2 = B,X 3 = R } + P { X 1 = R,X 2 = R,X 3 = B } = P { X 1 = B } P { X 2 = R | X 1 = B } P { X 3 = R | X 2 = R,X 1 = B } + P { X 1 = R } P { X 2 = B | X 1 = R } P { X 3 = R | X 2 = B,X 1 = R } + P { X 1 = R } P { X 2 = R | X 1 = R } P { X 3 = B | X 2 = R,X 1 = R } = 1 2 × 1 3 × 2 4 + 1 2 × 1 3 × 2 4 + 1 2 × 2 3 × 1 4 = 1 4 . 2. Uniform arrival. The arrival time of a professor to his office is uniformly distributed in the interval between 8 and 9 am. Find the probability that the professor will arrive during the next minute given that he has not arrived by 8:30. Repeat for 8:50. Solution: Let A be the event that ”the professor arrives between 8:30 and 8:31”, and let B be the event that ”the professor does not arrive before 8:30”. Then we have P ( A | B ) = P ( A ∩ B ) P ( B ) = P ( A ) P ( B ) = 1 / 60 1 / 2 = 1 30 . Similarly, P ( { Professor arrives between 8:50 and 8:51 }|{ He does not arrive before 8:50 } ) = 1 10 . 3. Probabilities from a cdf. The cdf of the random variable X is given by F X ( x ) = 1 , x > , 1 3 + 2 3 ( x + 1) 2 , − 1 ≤ x ≤ , , x < − 1 ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 14

hw2sol - UCSD ECE 153 Handout#11 Prof Young-Han Kim...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online