Question

Problem 6. Consider the n independent trails in Problem 5. Let On be the probability that there is no three consecutive successes in n trails. (1). Show that limn+cQn = 0 (2). Show that Qn = (1 - pQn-1 + p(1 - pQn-2 + p (1 - p)Qn-3 for n 3 (Hint: condition on the first failure).

Answer 1

5) Here $P_n$ is the probability that we get even number of successes in trials.

1)This can happen in 2 ways.

i) Odd number of success in first trials followed by a success whose probability is the compound probability,

T-U

pl1 - Pn-1)

ii) Even number of success in first trials followed by a failure whose probability is the compound probability,

T-U

(1 - pPn-1

The above cases being disjoint (probabilities can be added),

Pn = p 1 - Pn-1) + (1 - pPn-1

The proof is complete.

2) The above equation can be rewritten as

Pn = p + (1 - 2pPn-1 Pn = p + (1 - 2p) (p + (1 - 2pPn-2) Pn = p + (1 - 2pp + (1 - 2p) Pn-2 Pn = p + (1 – 2p)p+ (1 – 2p) (p + (1 - 2pPn-3) Pn = p + (1 - 2p)p + (1 - 2p)?p+ (1 - 2p)3p + ... + (1 - 2p)"-1p+ (1 - 2p)"P.

Note that (0 is even and always happens). The sum to above geometric series is

Po = 1

Pn=p+(1 – 2p)p+(1-2p)2p+ (1 - 2p)p+ ... + (1 - 2p)"-1p+ (1 – 2p)" 1-(1 - 2p)" - + (1 - 2p)" _1+ (1 - 2p)" Pn = p 2p

The limit is (note that )

11 – 2p <1

n+00 1 + (1 - 2p lim Pn = lim 1 + limno(1 – 2p)" P, = = 2. n 00 lim Jim P, 39 lim Pn =

Thus, it is proved that
lim P no"
does not depend on .