In a Little League baseball game, team A’s pitcher throws a strike 50% of the time and a ball 50% of the time, successive pitches are independent of one another, and the pitcher never hits a batter. Knowing this, team B’s manager has instructed the first batter not to swing at anything. Calculate the probability that a. The batter walks on the fourth pitch b. The batter walks on the sixth pitch (so two of the first five must be strikes), using a counting argument or constructing a tree diagram c. The batter walks d. The first batter up scores while no one is out (assuming that each batter pursues a no-swing strategy)

Solution : Step 1: It is given that in a Little League baseball game, team A’s pitcher throws a strike 50% of the time and a ball 50% of the time. Also given that successive pitches are independent of one another, and the pitcher never hits a batter. The team B manager instructed to the first batter not swing at anything. Step 2 : a) We have to find the probability that the batter walks on 4th pitch. Since it is given that a pitcher throws a strike 50% of the time and a ball 50% of the time. So there is an equal chances for both. Means probability of a pitcher throws a strike = probability that a batter throws a ball = 0.50= ½ So The P( the batter walks on the 4th pitch)= (½)^4 = .0625 b) we have to find the probability that the batter walks on the sixth pitch. Which means two of the first five must be strikes. P( the batter walks on the 6th pitch)=P((3 balls, 2 strike)+ball) Since the successive pitches are independent =P( 3 balls, 2 strikes)× P(ball) = 5 choose 2 × (½)^5 × (½) 5 = C2 × (½)^5 ×(½) = 10×(1/64) = 0.15625