Can someone please explain how to do this problem?
Woodrow wins a tic-tac-toe game 65% of the time when he chooses the first square and 32% of the time when his opponent chooses the first square. The player who plays first is chosen by a coin toss. What is the probability that Woodrow wins a given game?
I think you subtract or add 32% to or from 65% but I'm not sure.
.5 * .65 + .5 *.32 = .325 + .16 = .485
check
probability other player wins
= .5 * .35 + .5*.68 = .515
sum = 1, good :)
To find the probability that Woodrow wins a given game, we need to consider two cases: when Woodrow chooses the first square and when his opponent chooses the first square.
Case 1: Woodrow chooses the first square
According to the problem, Woodrow wins 65% of the time when he chooses the first square. So the probability of Woodrow winning under this case is 65%.
Case 2: Opponent chooses the first square
In this case, Woodrow wins 32% of the time. However, we need to consider that Woodrow doesn't always get to choose the first square. The player who goes first is chosen by a coin toss, which means that Woodrow goes first 50% of the time and his opponent goes first 50% of the time. So the probability of Woodrow winning when his opponent chooses the first square would be 50% multiplied by 32%, which equals 16%.
To calculate the overall probability that Woodrow wins a given game, we need to find the combined probability of both cases. Since the two cases are mutually exclusive, we can simply add the probabilities together.
Overall probability = (probability of Case 1) + (probability of Case 2)
= 65% + 16%
= 81%
Therefore, the probability that Woodrow wins a given game is 81%.
Pr(winning) = sum of prob of the ways to win...
Pr= 1/2 * .65 + 1/2 * .32= you do it.