Massachusetts Institute of Technology

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1 . (a) Massachusetts Institute of Technology Department of Electrical Engineering & omputer Science 6.0/6.: Probabilistic Systems nalysis (Fall 00) Problem Set : Solutions Due: September, 00 (b) ( c c ) ( c c ) ( c c ) ( c c c ) (c) ( ) c = c c c (d) (e) ( c c ) ( c c ) ( c c ) (f) c (g) ( c c ) (a) (b) (c) (d) (e) (f) (g). Since all outcomes are equally likely we apply the discrete uniform probability law to solve the problem. To solve for any event we simply count the number of elements in the event and divide by the total number of elements in the sample space. There are possible outcomes for each flip, and flips. Thus there are = 8 elements (or sequences) in the sample space. (a) ny sequence has probability of /8. Therefore P({H, H, H}) = /8. (b) This is still a single sequence, thus P({H, T, H}) = /8. (c) The event of interest has unique sequences, thus P({HHT, HTH, THH}) = /8. (d) The sequences where there are more heads than tails are : {HHH, HHT, HTH, THH}. unique sequences gives us P() = /.. The easiest way to solve this problem is to make a table of some sort, similar to the one below.

2 Department of Electrical Engineering & omputer Science 6.0/6.: Probabilistic Systems nalysis (Fall 00) Die Die Sum Total P(Sum) p p p p p p p 6p p p 6p 7p p 6p 7p 8p 80p and therefore (a) P(ll outcomes) = 80p (Total from the table) p = 80 P(Even sum) = p + p + p + 6p + p + 6p + 6p + 8p = 0p = / (b) P(Rolling a and a ) = P(,) + P(,) = p + p = 0p = /8. P() The shaded area in the following figure is the union of lice s pick being greater than / and ob s pick being greater than /. ob / 0 / lice

3 Department of Electrical Engineering & omputer Science 6.0/6.: Probabilistic Systems nalysis (Fall 00) P() = P(both numbers are smaller than /) area of small square = total sample area (/)(/) = = = /6 6 P() In the following figure, the diagonal line represents the set of points where the two selected numbers are equal. ob / 0 / lice The line has an area of 0. Thus, area of line 0 P() = = = 0 total sample area P( D) Overlapping the diagrams we would get for P() and P(D), ob / / lice

4 Department of Electrical Engineering & omputer Science 6.0/6.: Probabilistic Systems nalysis (Fall 00) P( D) = double shaded area total sample area = (/)(/)(/) + (/)(/)(/) /8 + 6/8 = = /7. (a) The probability of Mike scoring 0 points is proportional to the area of the inner disk. Hence, it is equal to απr = απ, where α is a constant to be determined. Since the probability of landing the dart on the board is equal to one, απ0 =, which implies that α = /(00π). Therefore, the probability that Mike scores 0 points is equal to π/(00π) = 0.0 (b) In order to score exactly 0 points, Mike needs to place the dart between and inches from the origin. n easy way to compute this probability is to look first at that of scoring more than 0 points, which is equal to απ = Next, since the 0 points ring is disjoint from the 0 points disc, probability of scoring more than 0 points is equal to the probability of scoring 0 points plus that of scoring exactly 0 points. Hence, the probability of Mike scoring exactly 0 points is equal to = 0.08 (c) For the part (a) question. The probability of John scoring 0 points is equal to the probability of throwing in the right half of the board and scoring 0 points plus that of throwing in the left half and scoring 0 points. The first term in the sum is proportional to the area of the right half of the inner disk and is equal to απr / = απ/, where α is a constant to be determined. Similarly, the probability of him throwing in the left half of the board and scoring 0 points is equal to βπ/, where β is a constant (not necessarily equal to α). In order to determine α and β, let us compute the probability of throwing the dart in the right half of the board. This probability is equal to απr / = απ0 / = α0π. Since that probability is equal to /, α = /(7π). In a similar fashion, β can be determined to be /(0π). onsequently, the total probability is equal to /0 + /00 = 0.0 For the part (b), The probability of scoring exactly 0 points is equal to that of scoring more than 0 points minus that of scoring exactly 0. y applying the same type of analysis as in (b) above, the probability is found to be equal to 0.08 These numbers suggest that John and Mike have similar skills, and are equally likely to win the game. The fact that Mike s better control (or worst, depending on how you look at it) of the direction of his throw does not increase his chances of winning can be explained by the observation that both players control over the distance from the origin is identical. 6. See the textbook, Problem. page, which proves the general version of onferroni s inequality. G. (a) If we define n = [a n, b n ] for all n, it is easy to see that the sequence,,... is monotonically decreasing, i.e., n+ n for all n:

5 Department of Electrical Engineering & omputer Science 6.0/6.: Probabilistic Systems nalysis (Fall 00) Omega... n Furthermore, n n = [a, b]. y the continuity property of probabilities (see Problem., page 6 of the text), lim P([a n, b n ]) = P([a, b]). n (b) No. onsider the following example. Let a n = a + n, b n = b n for all n. Then {a n } is a decreasing sequence that converges to a, and {b n } is an increasing sequence that converges to b. If we define a probability law that places non-zero probability only on points a and b, then lim n P([a n, b n ]) = 0, but P([a, b]) =. This example is closely related to the continuity property of probabilities. In this case, if we define n = [a n, b n ], then,,... is monotonically increasing, i.e., n n+, but = ( n n ) = (a, b), which is an open interval whose probability is 0 under our probability law.

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