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Unformatted text preview: EE 131A Problem Set #4 Probabilities Instructor: Vwani Roychowdhury 1. Problem 2.42 We assume (reasonably) that each time we try to catch an animal this is as likely to be caught as any of the remaining uncaught animals. The number of ways we can capture 5 tagged animals out of the 10 animals previously tagged is ‡ 10 5 · . The other 205 = 15 animals are untagged (belong to the remaining N 10 animals) and therefore can be captured in ‡ N 10 20 5 · ways. It follows that the probability that 5 animals out of 20 are found to be tagged if we have a population of N animals of which 10 were previously tagged is: ‡ 10 5 ·‡ N 10 20 5 · ‡ N 20 · = P 5 ( N ) Since P 5 ( N ) represents the probability of the observed event when there are actually N animals present in the region, it would appear that a reasonable estimate of N would be the value of N that maximizes P 5 ( N ). Such an estimate is called a maximumlikelihood estimate. The maximization of P 5 ( N ) can most simply be done by first noting that P 5 ( N ) P 5 ( N 1) = ( N 10)( N 20) N ( N 10 20+5) Now the above ratio is greater than 1 if and only if ( N 10)( N 20) ≥ N ( N 10 20+5) or, equivalently, if and only if N ≤ 40. Thus, as N increases, P 5 ( N ) first increases, and then decreases, and reaches its maximum value at the greatest integral value not exceeding (10 × 20) / 5 = 40. (In this case we happen to get an integral value). Hence, the maximumlikelihood estimate of N is 40. 2. Problem 2.45 We need to solve the equation x 1 + x 2 + x 3 = 7 , with the restrictions that 1 ≥ x i ≥ 6. If we rewrite y i = x i + 1, now with 0 ≥ y i ≥ 5 we have: y 1 + y 2 + y 3 = 4. The upper bound is satisfied automatically. If we view the above equation as four 1s that must be distributed in 4 distinct slots, we can see that the problem is equivalent 1 with the problem of choosing 4 out of 3 distinct objects with repetition allowed. Thus, the number of ways that 3 tosses of die can sum to 7 is ‡ 3 1+4 4 · = 15. Hence, the required probability is P [ sum = 7] = 15 6 3 = 0 . 0694 3. Problem 2.93 (a) The board will be working if only one of the 8 chips fails (i.e. 7 out of 8 chips work) or if all 8 of them work (in which case the backup chip wont be used).work) or if all 8 of them work (in which case the backup chip wont be used)....
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 Spring '05
 Roychowdhary
 Probability, Trigraph, p1, Genotypephenotype distinction, gene pair xx

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