10ProbStatHW3Sol

# 10ProbStatHW3Sol - SIEO 3600.002 Introduction to...

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

SIEO 3600.002 Solutions Assignment #3 Introduction to Probability and Statistics February 7, 2010 Prof. Chenxin Li Page 1 of 5 Solutions Assignment #3 1. A group of 5 boys and 10 girls is lined up in random order - that is, each of the 15! is assumed to be equally likely. (a) What is the probability that the person in the 4th position is a boy? To calculate this probability ﬁrst we need to calculate in how many diﬀerent ways we can organize the 15 diﬀerent kids in the line, regardless of their gender, which is equal to the permutation of 15 diﬀerent objects (15!). We know that position 4 in the line is gonna be a boy, then we have 14 positions left to organize the other 14 kids (regardless of their gender), Then the total number of diﬀerent ways that we can organize the other 14 kids is the permutation of 14 diﬀerent objects (14!). Finally let denote the event gender of kid in position 4 as p 4 , and this event has to possible outcomes b(boy) or g(girls). Then the probability that a boy (any one of the 5 boys) is in position 4, is: P ( p 4 = boy ) = 14! 15! · 5 = 1 3 (b) What about the person in the 12th position? To answer this question you should realize that the probability that the kid in position i is a boy is independent of i . Then this probability is exactly the same that the calculate in (a). P ( p 12 = boy ) = 14! 15! · 5 = 1 3 (c) What is the probability that a particular boy is in the 3rd position? If we are speaking about the speciﬁc boy j ( j goes from 1 to 5), then we just need to consider the total number of ways to organize the 15 kids in a line, and the total number of ways to organize 14 kids in 14 empty positions given that boy j is in position 3. P ( p 3 = boy j ) = 14! 15! = 1 15 2. A woman has n keys, of which one will open her door. If she tries the keys at random, discarding those that do not work, what is the probability that she will open the door on her k th try? What if she does not discard previously tried keys? case 1 The woman discard the keys that are not working. Let the K denote the time she gets the right key. Then, P ( K = 1) = 1 n , and notice that P ( K = 2 | K 6 = 1) = 1 n - 1 . Using conditional probability we have: P ( K = 2) = P ( K = 2 | K 6 = 1) · P ( K 6 = 1) = 1 n - 1 · ± 1 - 1 n ² = 1 n

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

View Full Document
2 SIEO 3600, Solutions Assignment #3 Continuing in the same fashion we have:
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 5

10ProbStatHW3Sol - SIEO 3600.002 Introduction to...

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

View Full Document
Ask a homework question - tutors are online