Solution Found!
Suppose that n independent trials are performed, with
Chapter 3, Problem 17TE(choose chapter or problem)
Problem 17TE
Suppose that n independent trials are performed, with trial i being a success with probability 1/(2i+ 1). Let Pndenote the probability that the total number of successes that result is an odd number.
(a) Find Pnfor n =1, 2, 3, 4, 5.
(b) Conjecture a general formula for Pn.
(c) Derive a formula for Pnin terms of Pn−1
(d) Verify that your conjecture in part (b) satisfies the recursive formula in part (c). Because the recursive formula has a unique solution, this then proves that your conjecture is correct.
Questions & Answers
QUESTION:
Problem 17TE
Suppose that n independent trials are performed, with trial i being a success with probability 1/(2i+ 1). Let Pndenote the probability that the total number of successes that result is an odd number.
(a) Find Pnfor n =1, 2, 3, 4, 5.
(b) Conjecture a general formula for Pn.
(c) Derive a formula for Pnin terms of Pn−1
(d) Verify that your conjecture in part (b) satisfies the recursive formula in part (c). Because the recursive formula has a unique solution, this then proves that your conjecture is correct.
ANSWER:
Step 1 of 4
(a)
Suppose that independent trials are performed, with trial
being a success with probability
Let
denote the probability that the total number of successes that result is an odd number.
We are asked to find the for
We have given trial being a success with probability,
……..(1)
Let be the event the
trial results in success and let
be the event the number of successes in
total trials is odd.
The probability is the probability the first trial is a success so we have
The probability is the union of the independent events where the first trial is a success and the second trial is a failure or the first trial is a failure and the second trial is a success.
[using equation (1)]
The probability is the union of the independent events where the three total success or one success and two failures.
In the same way the probability is the union of the independent events where the three total success with one failure and one success with three failure
\(P\left(E_{4}\right)=P\left(S_{1}^{c} S_{2} S_{3} S_{4}\right)+P\left(S_{1} S_{2}^{c} S_{3} S_{4}\right)+P\left(S_{1} S_{2} S_{3}^{c} S_{4}\right)+P\left(S_{1} S_{2} S_{3} S_{4}^{c}\right)+P\left(S_{1}^{c} S_{2}^{c} S_{3}^{c} S_{4}\right)\)
Here we can see the associated pattern with this probability which we can write,
………(2)
Hence,