Answer to Question #287085 in Statistics and Probability for Hlaidman

Question

Larry and Curly decide to have a chess match. The probability of Larry winning is 0.4. Each time Larry wins, he gains confidence and the probability of winning the next game increases by 0.1. Each time he loses, his confidence goes down and his probability of winning the next game decreases by 0.05. Determine the probability that in a three game match Larry wins at least two games.

Answer 1

Probability of larry winning = 0.4

The probability of larry winning is increased by 0.1 when he won the previous game

Probability of larry winning decrease by 0.05 when he lost the previous game

Let X represent the no of matches won by larry

Now we have to find probability that larry win atleast 2 matches

P(X≥2) = P(X=2)+ P(X=3)

Now P(X=2) in this probability we have 3 cases

Case 1= win, win, lose

Probability of case 1

$= 0.4 \times (0.4+0.1)\times (1-(0.4+0.1+0.1)) \\ =0.4 \times 0.5 \times 0.4 \\ =0.08$

Case 2=win, lose ,win

Probability of case 2

$=0.4 \times (1-(0.4+0.1)) \times (0.5-0.05) \\ =0.4 \times 0.5 \times 0.45 \\ =0.09$

Case 3 = lose, win, win

Probability of case 3

$=(1-0.4) \times (0.4-0.05) \times (0.35+0.1) \\ =0.6 \times 0.35 \times 0.45 \\ =0.0945 \\ P(X=2) = case1+case2+case3 \\ =0.08+0.09+0.0945 \\ =0.2645 \\ P(X=3) = 0.4 \times (0.4+0.1) \times (0.5+0.1) \\ = 0.4 \times 0.5 \times 0.6 \\ =0.12 \\ P(X≥2)=0.2645+0.12=0.3845$