Unformatted text preview: EE 131A Homework #2 Solution Winter 2007 K. Yao 1. There are 4 × 3 × 3 = 36 diﬀerent designs. 2. The number of distinct ordered triplets with replacement equals 7 × 2 × 52 = 728. 3. On the ﬁrst day, there is a choice of 5 pairs, and subsequently, there is a choice of only 4 pairs. Total number is = 5 × 4 × 4 × 4 × 4 × 4 = 5 × 45 = 5120. 4. There are 10 choices for plant A, but only 9 for plant B, and 8 for plant C. This give 10 a total of P3 = 10!/7! = 10 × 9 × 8 = 720 ways of assigning employees to the plants. 5. The number of ordering is the permutation of four things take 4 at a time. Thus, 4 P4 = 4!/0! = 4! = 4 × 3 × 2 × 1 = 24. 6. Here order is not important and we only want to know how many subsets of size 3 can 10 be selected from 10 people. Thus, C3 = 10!/(3!7!) = (10 × 9 × 8)/(1 × 2 × 3) = 120.
10 7. We already know there are C3 = 120 ways to select 3 employees from 10. Similarly, 2 8 there are C1 = 2 ways of selecting 1 woman from the 2 available and C2 = 28 ways to select 2 men from the 8 available. If selections are at random, then the probability of 28 10 selecting just one woman and two men is P = C1 C2 /C3 = 2×28/120 = 7/15 = 0.4667. 100 8. Combinations of M out of 100 equals CM . Number of ways of choosing m defeck 100 tives out of k and M − m nondefectives out of 100 − k equals Cm CM −−k . Thus, m k 100−k 100 P {m defectives in M samples} = Cm CM −m /CM . 9 9. There are C6 ways of getting 6 numbers from the 9 numbers without replacement. 3 5 There are C2 ways of picking 2 numbers out of 17, 21, and 28, while there are C3 ways of picking 3 numbers from 2,3, 7, 8, and 12. The probability the third largest number 35 9 is 15 is then P = C2 C3 /C6 = 5/14 = 0.3571. ...
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 Spring '08
 LORENZELLI
 Harshad number, C2 C3 /C6, Cm CM −−k, Cm CM −m

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