(A touchdown is worth 6 points. An extra point is worth 1 point. A two-point conversion is worth 2 points. After a team scores a touchdown, they can either attempt an extra point or attempt a two-point conversion. The probability that an extra point attempt is converted is 0.98. The probability that a two-point conversion attempt is converted is 0.44. All extra point attempts and two-point conversion attempts are independent.) The Bears are down by 14 points late in the fourth quarter to the Packers. Let's assume that the only way the Bears can Win is to score 14 points (via two touchdowns and two extra points or two touchdowns and one two-point conversion) and win in overtime, or to score 15 points (via two touchdowns, an extra point, and a twopoint conversion) and win in regulation. Let's stipulate that the Bears do score the required two touchdowns in the nal minutes and do not give up any more points to the Packers, and let's assume that each team has an equal chance of winning in overtime should the game go that far. Given these stipulations and assumptions, what is the probability that the Bears will win if they pursue the following strategy: go for two after the rst touchdown and then only go for two after the second touchdown if the rst two-point conversion attempt fails. (Round your nal answer to four decimal places if necessary.)