I am going to take my chances and answer this in the morning if nobody else does. I have some bad phys. ed. memories that I don't want to become nightmares.😀 Good night everyone.

• The chances of Jancgarlest getting points is J: 0.45
• Chrisrice not getting the point is C: 1.00-0.18 = 0.82...
• Chances of both events happening is the product of J & C: 0.82*0.45 = 0.369

The chances of Jan scoring but not Chris after they both throw is %36.9

Posted via D.Buzz

Forgot about the time. Someone else already gave the correct answer of 36.9% but I'll show it my way.
Let X = P(A and Not B) be the probability of Jan and not Chris
Let P(A) = Probability of Jan = 0.45
Let P(B) = Probability of Chris = 0.18

1. P(A AND B) = (0.18)(0.45) = 0.081
2. X = P(A) - P(A AND B) = 0.45 - 0.081= 0.369 = 36.9%

Answer for Day 6 Math Problem

36.9% 🎯

The problem is basically the chance of @jancharlest shooting the ball multiplied by the chance of @chrisrice not shooting the ball. The solution is 82% multiplied by 45% (or 0.45), whose result is 36.9%.

Winner: @ahmadmangazap 🏅

I don't know how the percent (%) symbol is written in @ahmadmangazap's country, but his answer is clear.

1 HIVE has been sent to @ahmadmangazap's account as reward. 💰

Mentions: @holovision and @jfang003
Special mentions: @chrisrice and @jancharlest 😆