Justin ,cam and Ben are playing a board game where exactly one player will win .Ben estimates that Justin has a 20% chance of winning each game and that cam has a 50% chance of winning each game?

Question :

Justin, Cam, and Ben are playing a board game where exactly one player will win. Ben estimates that Justin has a 20%, percent chance of winning each game and that Cam has a 50%, percent chance of winning each game. What is the probability that Ben will win the board game?

30%

Step-by-step explanation:

Given the following :

Justin's chance of winning each game = 20%

Cam's chance of winning each game = 50%

Therefore, probability of :

Justin winning = 20/100 = 0.2

Cam winning = 50/100 = 0.5

Overall probability = 100% = 1

Therefore,

P(Justin winning) + P(Cam winning) + P(Ben winning) = 1

0.2 + 0.5 + P(Ben winning) = 1

P(Ben winning) = 1 - 0.7

P(Ben winning) = 0.3

0.3 = 30%

3,000,540

step-by-step explanation:

there are 70 more legs than eyes at the dog show (196 – 126 = 70). dogs have two more legs than eyes, and humans have the same number of legs as eyes. thus 70 is twice the number of dogs at the show, so the number of dogs is half of 70, which is 35. to get the number of humans, subtract 35 from 63: 63 – 35 = 28.

step-by-step explanation: