midterm2sol

By the same reasoning pren en1 is the probability that

Info iconThis preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: the probability that the n-th flip was tails; this is because if we had an even number of heads on day n − 1, we must obtain tails on day n to maintain an even number of heads. By ¯ the same reasoning, Pr[En |En−1 ] is the probability that the n-th flip was heads. Plugging in the values and the inductive hypothesis we obtain: 1 (1 − 2 p)n−1 1 (1 − 2 p)n−1 Pr[En ] = (1 − p)( + ) + p(1 − ( + )). 2 2 2 2 After simplification, we find that: Pr[En ] = 1 (1 − 2 p)n + . 2 2 Here is another way to think about this problem. To get an even number of heads in n tosses, you must have either an even number of heads in n − 1 tosses (event En−1 ) followed by a tails in the nth ¯ toss (event T ) or an odd number of heads in n − 1 tosses (event En−1 ) followed by a heads in the nth toss (event H ). Note that these two possibilities are disjoint, so the total probability is the sum of the probabilities of each of the two. It follows that: ¯ Pr[En ] = Pr[En−1 ] Pr[T ] + Pr[En−1 ] Pr[H ] Here we used the fact that the outcome of the n-th coin toss is independent of the outcomes of the first n − 1 tosses. We apply the induction hypothesis as above to get: 1 (1 − 2 p)n−1 1 (1 − 2 p)n−1 1 (1 − 2 p)n Pr[En ] = (1 − p)( + ) + p(1 − ( + )) = Pr[En ] = + . 2 2 2 2 2 2 CS 70, Fall...
View Full Document

This document was uploaded on 01/28/2014.

Ask a homework question - tutors are online