Probability problem about trials

Question

Independent trials are performed, each with probability p of success. Let P_n be the probability that n trials result in an even number of successes. Show that:

Answer 1

Denote P_n(even) as a probability that n trials result an even number of successes.

P_n(odd) as a probability that n trials result an odd number of successes.

If P_n(even) = 0.5[1 + (1-2p)ⁿ], then

Prove these formulas using mathematical induction.

Step 1. Let n = 1.

Then P₁(even) = P(1 trial results 0 successes) = 1 - p

Verify the formula: P₁(even) = 0.5[1 + (1-2p)¹] = 0.5[2 - 2p] = 1 - p

The formula is confirmed.

Step k. Suppose for n = k P_k(even) = 0.5[1 + (1-2p)^k]

Then P_k(odd) = 1 - P_k(even) = 0.5[1 - (1-2p)^k]

Step k+1. Now let n = k+1

P_k+1(even) = P(even successes in k+1 trials) =

= P(even successes in k trials and failure for the last trial) + P(odd successes in k trials and success for the last trial) =

= P_k(even) (1-p) + P_k(odd) p = 0.5[1 + (1-2p)^k] (1-p) + 0.5[1 - (1-2p)^k] p =

= 0.5[1 - p + (1-2p)^k - (1-2p)^k p + p - (1-2p)^k p] = 0.5[1 + (1-2p)^k - (1-2p)^k 2p] = 0.5[1 + (1-2p)^k (1 - 2p)] =

= 0.5[1 + (1-2p)^k+1]

The formula is proved.