The probability that he is left with no money by the 14^th round or earlier is

Question

A player ‘A’ plays a game against a machine. At each round he deposits one rupee in a slot and then flips a coin which has a probability p of showing a head. If the flipped coin show head, he gets back the rupee he deposited and one more rupee from the machine, else he loses his rupee. Let A starts with 10 rupee coins and q = 1 – p (the probability of showing a tail), then

The probability that he is left with no money by the 14^th round or earlier is 

(A) q¹⁰ (1 + 10pq + 65p²q²) 

(B) q¹⁴ (p²q + 36pq + 7) 

(C) q¹² + 3pq¹³ + 3p¹³q +p¹² 

(D) 1 – ¹⁰C₁ pq₁₁ – ¹⁰C₂p²q¹² 

Answer 1

Answer is (A) q¹⁰ (1 + 10pq + 65p²q²)

To drain out at the 14^th round, two cases arise 

(i) He gets exactly 2 heads in the first 10 rounds. 

Therefore, probability in this case is 

¹⁰C₂ p²q⁸ . q⁴ = 45p²q¹² 

(ii) He gets exactly 1 head in the first 10 rounds and then exactly one head at the next two rounds. 

Therefore, probability in this case is 

¹⁰C₁ pq⁹ . 2 C₁ pq . q² 

= 20p² q¹² 

To drain out earlier than 14^th round, two cases arise 

(i) He gets no head in the first 10 rounds.

Therefore, probability in this case is q¹⁰.

(ii) He gets exactly one head in first 10 rounds and then no heads.

Therefore, probability in this case is 

¹⁰C₁ q⁹ . p. q² = 10pq¹¹

Therefore, 

required probability = 65 p²q¹² + q¹⁰ + 10pq¹¹